Topic Quiz • 150 questions • 25 seconds per question.
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What is the value of \( \frac{18!}{15!}-\frac{17!}{14!} \)?
B. (816)
पहला पद \(18\cdot17\cdot16=4896\) और दूसरा \(17\cdot16\cdot15=4080\) है। अंतर (816) है। / The first term is \(18\cdot17\cdot16=4896\) and the second is \(17\cdot16\cdot15=4080\). The difference is (816).
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What is the simplified value of \( \frac{15!}{11!\cdot4!} \)?
C. (1365)
\( \frac{15\cdot14\cdot13\cdot12}{4!}=1365 \) है। हर में (4!) का सही मान लगाएं। / \( \frac{15\cdot14\cdot13\cdot12}{4!}=1365 \). Use the correct value of (4!) in the denominator.
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If ( \frac{(n+4)!}{n!}=11880 ), what will be the value of (n)?
C. (8)
यह ((n+4)(n+3)(n+2)(n+1)=11880) है और \(12\cdot11\cdot10\cdot9=11880\)। इसलिए (n=8) है। / It is ((n+4)(n+3)(n+2)(n+1)=11880), and \(12\cdot11\cdot10\cdot9=11880\). So (n=8).
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If ( \frac{(n+2)!+(n+1)!}{(n+1)!}=14 ), what is the value of (n)?
C. (11)
सरल रूप (n+3) है। इसलिए (n+3=14) और (n=11)। / The simplified form is (n+3). Thus (n+3=14) and (n=11).
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What is the simplified form of ( \frac{(n+3)!-(n+2)!}{(n+1)!} )?
A. ((n+2)2)
((n+3)!-(n+2)!=(n+2)!((n+3)-1)) है। इससे ((n+2)2) मिलता है। / ((n+3)!-(n+2)!=(n+2)!((n+3)-1)). This gives ((n+2)2).
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If ( \frac{(n+3)!-(n+2)!}{(n+1)!}=121 ), what will be the value of (n)?
B. (9)
सरल रूप ((n+2)2) है। इसलिए (n+2=11) और (n=9)। / The simplified form is ((n+2)2). Therefore (n+2=11) and (n=9).
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If ( \frac{(n!)2}{(n-2)!(n+2)!}=\frac{10}{21} ), what is the value of (n)?
B. (5)
सरल रूप ( \frac{n(n-1)}{(n+1)(n+2)} ) है। (n=5) रखने पर \( \frac{20}{42}=\frac{10}{21} \) मिलता है। / The simplified form is ( \frac{n(n-1)}{(n+1)(n+2)} ). Putting (n=5) gives \( \frac{20}{42}=\frac{10}{21} \).
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What is the highest power of (2) that divides (20!)?
D. (18)
घात \( \left\lfloor\frac{20}{2}\right\rfloor+\left\lfloor\frac{20}{4}\right\rfloor+\left\lfloor\frac{20}{8}\right\rfloor+\left\lfloor\frac{20}{16}\right\rfloor=18 \) है। अभाज्य घात के लिए सभी भागफल जोड़ें। / The exponent is \( \left\lfloor\frac{20}{2}\right\rfloor+\left\lfloor\frac{20}{4}\right\rfloor+\left\lfloor\frac{20}{8}\right\rfloor+\left\lfloor\frac{20}{16}\right\rfloor=18 \). Add all quotients for a prime exponent.
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How many zeros will be at the end of (45!)?
A. (10)
शून्यों की संख्या \( \left\lfloor\frac{45}{5}\right\rfloor+\left\lfloor\frac{45}{25}\right\rfloor=10 \) है। (25) के योगदान को न भूलें। / The number of zeros is \( \left\lfloor\frac{45}{5}\right\rfloor+\left\lfloor\frac{45}{25}\right\rfloor=10 \). Do not forget the contribution of (25).
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What is the smallest positive (n) for which (n!) is divisible by (1440)?
C. (8)
\(1440=2^5\cdot3^2\cdot5\) है और यह जरूरत पहली बार (8!) में पूरी होती है। विभाज्यता में अभाज्य गुणनखंड जांचें। / \(1440=2^5\cdot3^2\cdot5\), and this requirement is first satisfied by (8!). For divisibility, check prime factors.
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What is the value of \( \frac{20!}{18!\cdot2!}+\frac{19!}{17!\cdot2!} \)?
C. (361)
दोनों पद (190) और (171) हैं। योग (361) है। / The two terms are (190) and (171). Their sum is (361).
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What is the value of \( \frac{21!}{18!\cdot3!}-\frac{20!}{17!\cdot3!} \)?
C. (190)
पहला पद (1330) और दूसरा (1140) है। अंतर (190) है। / The first term is (1330) and the second is (1140). The difference is (190).
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What is the simplified value of \( \frac{14!}{10!\cdot4!}\div\frac{13!}{9!\cdot4!} \)?
B. \(\frac{7}{5}\)
दोनों पद (1001) और (715) हैं। अनुपात \( \frac{1001}{715}=\frac{7}{5} \) है। / The two terms are (1001) and (715). The ratio is \( \frac{1001}{715}=\frac{7}{5} \).
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Which is the correct expansion of ( \frac{(2n)!}{(2n-4)!} )?
B. ((2n)(2n-1)(2n-2)(2n-3))
((2n)!) को ((2n)(2n-1)(2n-2)(2n-3)(2n-4)!) तक फैलाते हैं। इसलिए चार गुणक बचते हैं। / We expand ((2n)!) up to ((2n)(2n-1)(2n-2)(2n-3)(2n-4)!). Therefore four factors remain.
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If ( \frac{(2n)!}{(2n-4)!}=1680 ), what is the value of (n)?
B. (4)
यह ((2n)(2n-1)(2n-2)(2n-3)=1680) है। \(8\cdot7\cdot6\cdot5=1680\), इसलिए (n=4)। / It is ((2n)(2n-1)(2n-2)(2n-3)=1680). Since \(8\cdot7\cdot6\cdot5=1680\), (n=4).
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What is ( \frac{(3n+2)!}{(3n)!} ) equal to?
A. ((3n+2)(3n+1))
((3n+2)!=(3n+2)(3n+1)(3n)!) है। इसलिए दो गुणक बचते हैं। / ((3n+2)!=(3n+2)(3n+1)(3n)!). Therefore two factors remain.
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If ( \frac{(3n+2)!}{(3n)!}=182 ), what is the value of (n)?
B. (4)
यह ((3n+2)(3n+1)=182) है और \(14\cdot13=182\)। इसलिए (3n+2=14) और (n=4)। / It is ((3n+2)(3n+1)=182), and \(14\cdot13=182\). Thus (3n+2=14) and (n=4).
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If ( \frac{(n+4)!}{(n+2)!}-\frac{(n+3)!}{(n+1)!}=24 ), what is the value of (n)?
C. (9)
सरल रूप (2(n+3)) है। इसलिए (2(n+3)=24) और (n=9)। / The simplified form is (2(n+3)). Thus (2(n+3)=24) and (n=9).
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What is the simplified form of ( \frac{(n+5)!+(n+4)!}{(n+3)!} )?
A. ((n+4)(n+6))
ऊपर ((n+4)!((n+5)+1)) बनता है। भाग देने पर ((n+4)(n+6)) मिलता है। / The numerator becomes ((n+4)!((n+5)+1)). Dividing gives ((n+4)(n+6)).
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If ( \frac{(n+5)!+(n+4)!}{(n+3)!}=120 ), what is the value of (n)?
B. (6)
सरल रूप ((n+4)(n+6)) है। (n=6) रखने पर \(10\cdot12=120\) मिलता है। / The simplified form is ((n+4)(n+6)). Putting (n=6) gives \(10\cdot12=120\).
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What is the value of \( \frac{12!}{8!\cdot4!}+\frac{12!}{9!\cdot3!} \)?
B. (715)
दोनों पद (495) और (220) हैं। योग (715) है। / The two terms are (495) and (220). Their sum is (715).
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What is the value of \( \frac{10!}{5!\cdot5!}\times\frac{5}{9} \)?
C. (140)
पहला भाग (252) है। \(252\cdot\frac{5}{9}=140\) मिलता है। / The first part is (252). Then \(252\cdot\frac{5}{9}=140\).
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What is the value of \( \frac{13!}{10!}-2\cdot\frac{12!}{10!} \)?
B. (1452)
पहला पद (1716) और दूसरा \(2\cdot132=264\) है। अंतर (1452) है। / The first term is (1716) and the second is \(2\cdot132=264\). The difference is (1452).
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What is the value of \( \frac{9!+2\cdot8!}{7!} \)?
C. (88)
\( \frac{9!}{7!}=72 \) और \( \frac{2\cdot8!}{7!}=16 \) है। कुल (88) मिलता है। / \( \frac{9!}{7!}=72 \) and \( \frac{2\cdot8!}{7!}=16 \). The total is (88).
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What is the simplified form of ( \frac{(n+3)!}{(n-1)!(n+2)(n+1)} )?
A. (n(n+3))
ऊपर ((n+3)(n+2)(n+1)n(n-1)!) है। काटने पर (n(n+3)) बचता है। / The numerator is ((n+3)(n+2)(n+1)n(n-1)!). After cancellation, (n(n+3)) remains.
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If ( \frac{(n+3)!}{(n-1)!(n+2)(n+1)}=108 ), what is the value of (n)?
C. (9)
सरल रूप (n(n+3)) है। \(9\cdot12=108\), इसलिए (n=9)। / The simplified form is (n(n+3)). Since \(9\cdot12=108\), (n=9).
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What is the highest power of (3) that divides (25!)?
C. (10)
घात \( \left\lfloor\frac{25}{3}\right\rfloor+\left\lfloor\frac{25}{9}\right\rfloor=10 \) है। (9) के योगदान को जोड़ना जरूरी है। / The exponent is \( \left\lfloor\frac{25}{3}\right\rfloor+\left\lfloor\frac{25}{9}\right\rfloor=10 \). Adding the contribution of (9) is necessary.
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What is the smallest positive (n) for which (n!) is divisible by \(2^7\cdot3^4\cdot5\)?
B. (9)
(9!) में (2) की घात (7) और (3) की घात (4) मिल जाती है। इससे छोटा (n) यह शर्त पूरी नहीं करता। / (9!) contains exponent (7) of (2) and exponent (4) of (3). No smaller (n) satisfies this condition.
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What is the value of \( \frac{25!}{23!\cdot2!}-\frac{24!}{22!\cdot2!} \)?
C. (24)
पहला पद (300) और दूसरा (276) है। अंतर (24) है। / The first term is (300) and the second is (276). The difference is (24).
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What is the value of \( \frac{16!}{12!\cdot4!}-\frac{15!}{12!\cdot3!} \)?
B. (1365)
पहला पद (1820) और दूसरा (455) है। अंतर (1365) है। / The first term is (1820) and the second is (455). The difference is (1365).
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Which is the correct expansion of ( \frac{(n+1)!}{(n-3)!} )?
A. ((n+1)n(n-1)(n-2))
((n+1)!) को ((n+1)n(n-1)(n-2)(n-3)!) तक फैलाते हैं। इसलिए चार गुणक बचते हैं। / We expand ((n+1)!) up to ((n+1)n(n-1)(n-2)(n-3)!). Therefore four factors remain.
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If ( \frac{(n+1)!}{(n-3)!}=5040 ), what is the value of (n)?
C. (9)
यह ((n+1)n(n-1)(n-2)=5040) है। \(10\cdot9\cdot8\cdot7=5040\), इसलिए (n=9)। / It is ((n+1)n(n-1)(n-2)=5040). Since \(10\cdot9\cdot8\cdot7=5040\), (n=9).
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What is the simplified form of ( \frac{\frac{(n+2)!}{(n-2)!}}{\frac{(n+1)!}{(n-3)!}} )?
A. \(\frac{n+2}{n-2}\)
समान गुणक ((n+1)n(n-1)) कट जाते हैं। इसलिए \( \frac{n+2}{n-2} \) बचता है। / The common factors ((n+1)n(n-1)) cancel out. Therefore \( \frac{n+2}{n-2} \) remains.
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If ( \frac{\frac{(n+2)!}{(n-2)!}}{\frac{(n+1)!}{(n-3)!}}=2 ), what is the value of (n)?
C. (6)
सरल रूप \( \frac{n+2}{n-2} \) है। \( \frac{n+2}{n-2}=2 \) से (n=6) मिलता है। / The simplified form is \( \frac{n+2}{n-2} \). From \( \frac{n+2}{n-2}=2 \), we get (n=6).
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How many zeros will be at the end of (30!)?
B. (7)
शून्यों की संख्या \( \left\lfloor\frac{30}{5}\right\rfloor+\left\lfloor\frac{30}{25}\right\rfloor=7 \) है। (25) का अतिरिक्त योगदान जोड़ें। / The number of zeros is \( \left\lfloor\frac{30}{5}\right\rfloor+\left\lfloor\frac{30}{25}\right\rfloor=7 \). Add the extra contribution of (25).
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What is the simplified value of \( \frac{22!}{20!}\div\frac{11!}{9!} \)?
B. \(\frac{21}{5}\)
मान \( \frac{22\cdot21}{11\cdot10}=\frac{21}{5} \) है। भाग में दोनों अनुपातों को फैलाकर काटें। / The value is \( \frac{22\cdot21}{11\cdot10}=\frac{21}{5} \). Expand and cancel both ratios in division.
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What is the simplified value of \( \frac{7!\cdot9!}{8!\cdot6!} \)?
B. (63)
\( \frac{7!}{6!}=7 \) और \( \frac{9!}{8!}=9 \) है। इसलिए मान (63) है। / \( \frac{7!}{6!}=7 \) and \( \frac{9!}{8!}=9 \). Hence the value is (63).
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What is the simplified form of ( \frac{(n+4)!-(n+3)!}{(n+2)!} )?
A. ((n+3)2)
((n+4)!-(n+3)!=(n+3)!((n+4)-1)) है। इसलिए ((n+3)2) मिलता है। / ((n+4)!-(n+3)!=(n+3)!((n+4)-1)). Therefore ((n+3)2) is obtained.
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If ( \frac{(n+4)!-(n+3)!}{(n+2)!}=144 ), what is the value of (n)?
B. (9)
सरल रूप ((n+3)2) है। इसलिए (n+3=12) और (n=9)। / The simplified form is ((n+3)2). Thus (n+3=12) and (n=9).
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What is the simplified form of ( \frac{(n+5)!}{(n+2)!}-\frac{(n+4)!}{(n+1)!} )?
B. (3(n+4)(n+3))
सामान्य ((n+4)(n+3)) निकालने पर अंतर ((n+5)-(n+2)=3) है। उत्तर (3(n+4)(n+3)) है। / Taking common ((n+4)(n+3)), the difference is ((n+5)-(n+2)=3). The answer is (3(n+4)(n+3)).
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If ( \frac{(n+5)!}{(n+2)!}-\frac{(n+4)!}{(n+1)!}=396 ), what is the value of (n)?
B. (8)
सरल रूप (3(n+4)(n+3)) है। \(3\cdot12\cdot11=396\), इसलिए (n=8)। / The simplified form is (3(n+4)(n+3)). Since \(3\cdot12\cdot11=396\), (n=8).
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If ( \frac{(n!)2}{(n-1)!(n+1)!}=\frac{11}{12} ), what is the value of (n)?
B. (11)
सरल रूप \( \frac{n}{n+1} \) है। \( \frac{n}{n+1}=\frac{11}{12} \) से (n=11) मिलता है। / The simplified form is \( \frac{n}{n+1} \). From \( \frac{n}{n+1}=\frac{11}{12} \), we get (n=11).
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What is the simplified value of \( \frac{18!}{14!\cdot4!}\div\frac{17!}{13!\cdot4!} \)?
A. \(\frac{153}{119}\)
दोनों पद (3060) और (2380) हैं। उनका अनुपात \( \frac{3060}{2380}=\frac{153}{119} \) है। / The two terms are (3060) and (2380). Their ratio is \( \frac{3060}{2380}=\frac{153}{119} \).
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What is the simplified form of ( \frac{(n+2)!}{n!}+\frac{n!}{(n-2)!} )?
B. \(2n^2+2n+2\)
पहला पद ((n+2)(n+1)) और दूसरा (n(n-1)) है। जोड़ने पर \(2n^2+2n+2\) मिलता है। / The first term is ((n+2)(n+1)) and the second is (n(n-1)). Adding gives \(2n^2+2n+2\).
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If ( \frac{(n+2)!}{n!}+\frac{n!}{(n-2)!}=114 ), what is the value of (n)?
B. (7)
सरल रूप \(2n^2+2n+2\) है। (n=7) रखने पर (98+14+2=114) मिलता है। / The simplified form is \(2n^2+2n+2\). Putting (n=7) gives (98+14+2=114).
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What is the value of \( \frac{12!}{6!\cdot6!}-\frac{11!}{5!\cdot6!} \)?
C. (462)
पहला पद (924) और दूसरा (462) है। अंतर (462) है। / The first term is (924) and the second is (462). The difference is (462).
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What is the simplified form of ( \frac{(n+6)!}{(n+2)!}-\frac{(n+5)!}{(n+1)!} )?
A. (4(n+5)(n+4)(n+3))
सामान्य ((n+5)(n+4)(n+3)) निकालने पर अंतर ((n+6)-(n+2)=4) है। पहले समान गुणकों को पहचानें। / Taking common ((n+5)(n+4)(n+3)), the difference is ((n+6)-(n+2)=4). Identify common factors first.
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If ( \frac{(n+6)!}{(n+2)!}-\frac{(n+5)!}{(n+1)!}=2880 ), what is the value of (n)?
B. (5)
सरल रूप (4(n+5)(n+4)(n+3)) है। \(4\cdot10\cdot9\cdot8=2880\), इसलिए (n=5)। / The simplified form is (4(n+5)(n+4)(n+3)). Since \(4\cdot10\cdot9\cdot8=2880\), (n=5).
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What is the smallest positive (n) for which (n!) is divisible by \(2^8\cdot3^2\cdot5^2\)?
C. (10)
(10!) में (2) की घात (8) और (5) की घात (2) मिलती है। (9!) में \(5^2\) नहीं आता, इसलिए (10) न्यूनतम है। / (10!) contains exponent (8) of (2) and exponent (2) of (5). (9!) does not contain \(5^2\), so (10) is minimum.
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What is the highest power of (7) that divides (100!)?
C. (16)
घात \( \left\lfloor\frac{100}{7}\right\rfloor+\left\lfloor\frac{100}{49}\right\rfloor=16 \) है। उच्च घातों जैसे (49) का योगदान जरूर जोड़ें। / The exponent is \( \left\lfloor\frac{100}{7}\right\rfloor+\left\lfloor\frac{100}{49}\right\rfloor=16 \). Always add the contribution of higher powers such as (49).
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What is the value of \( \frac{19!}{16!}-\frac{18!}{15!} \)?
B. (918)
पहला पद (5814) और दूसरा पद (4896) है। इसलिए अंतर (918) है। / The first term is (5814) and the second term is (4896). Therefore the difference is (918).
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What is the simplified value of \( \frac{16!}{12!\cdot4!} \)?
B. (1820)
(16!) को \(16\cdot15\cdot14\cdot13\cdot12!\) लिखें। (4!) से भाग देने पर मान (1820) मिलता है। / Write (16!) as \(16\cdot15\cdot14\cdot13\cdot12!\). Dividing by (4!) gives (1820).
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If ( \frac{(n+5)!}{(n+1)!}=11880 ), what is the value of (n)?
B. (7)
यह ((n+5)(n+4)(n+3)(n+2)=11880) देता है। \(12\cdot11\cdot10\cdot9=11880\), इसलिए (n=7)। / It gives ((n+5)(n+4)(n+3)(n+2)=11880). Since \(12\cdot11\cdot10\cdot9=11880\), (n=7).
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If ( \frac{(n+4)!+(n+3)!}{(n+2)!}=143 ), what will be the value of (n)?
B. (8)
सरल रूप ((n+3)(n+5)) है। \(11\cdot13=143\), इसलिए (n=8)। / The simplified form is ((n+3)(n+5)). Since \(11\cdot13=143\), (n=8).
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What is the simplified form of ( \frac{(n+5)!-(n+4)!}{(n+3)!} )?
A. ( (n+4)2 )
((n+5)!-(n+4)!=(n+4)!((n+5)-1)) है। इसलिए भाग देने पर ((n+4)2) मिलता है। / ((n+5)!-(n+4)!=(n+4)!((n+5)-1)). Therefore division gives ((n+4)2).
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If ( \frac{(n+5)!-(n+4)!}{(n+3)!}=169 ), what is the value of (n)?
B. (9)
अभिव्यक्ति ((n+4)2) बनती है। (n+4=13), इसलिए (n=9)। / The expression becomes ((n+4)2). Since (n+4=13), (n=9).
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What is the simplified form of ( \frac{(n!)2}{(n-3)!(n+3)!} )?
B. ( \frac{n(n-1)(n-2)}{(n+1)(n+2)(n+3)} )
( \frac{n!}{(n-3)!}=n(n-1)(n-2) ) और ( \frac{n!}{(n+3)!}=\frac{1}{(n+1)(n+2)(n+3)} )। दोनों भागों को गुणा करें। / ( \frac{n!}{(n-3)!}=n(n-1)(n-2) ) and ( \frac{n!}{(n+3)!}=\frac{1}{(n+1)(n+2)(n+3)} ). Multiply both parts.
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If ( \frac{(n!)2}{(n-3)!(n+3)!}=\frac{5}{21} ), what is the value of (n)?
B. (6)
सरल रूप में (n=6) रखने पर \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \) मिलता है। विकल्प जांचते समय कटौती करें। / Putting (n=6) in the simplified form gives \( \frac{6\cdot5\cdot4}{7\cdot8\cdot9}=\frac{5}{21} \). Cancel while checking options.
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What is the highest power of (2) that divides (50!)?
C. (47)
घात \( \left\lfloor\frac{50}{2}\right\rfloor+\left\lfloor\frac{50}{4}\right\rfloor+\left\lfloor\frac{50}{8}\right\rfloor+\left\lfloor\frac{50}{16}\right\rfloor+\left\lfloor\frac{50}{32}\right\rfloor=47 \) है। सभी भागफल जोड़ना जरूरी है। / The exponent is \( \left\lfloor\frac{50}{2}\right\rfloor+\left\lfloor\frac{50}{4}\right\rfloor+\left\lfloor\frac{50}{8}\right\rfloor+\left\lfloor\frac{50}{16}\right\rfloor+\left\lfloor\frac{50}{32}\right\rfloor=47 \). Adding all quotients is necessary.
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How many zeros will be at the end of (60!)?
B. (14)
शून्यों की संख्या \( \left\lfloor\frac{60}{5}\right\rfloor+\left\lfloor\frac{60}{25}\right\rfloor=14 \) है। (25) के अतिरिक्त योगदान को जरूर जोड़ें। / The number of zeros is \( \left\lfloor\frac{60}{5}\right\rfloor+\left\lfloor\frac{60}{25}\right\rfloor=14 \). Always add the extra contribution of (25).
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What is the smallest positive (n) for which (n!) is divisible by \(2^9\cdot3^3\cdot5^2\)?
C. (11)
(11!) में (2) की घात (8) नहीं बल्कि (8) से अधिक (8) ही होती है, पर \(2^9\) के लिए (12!) चाहिए। इसलिए सही न्यूनतम (12) है। / (11!) has exponent (8) of (2), so \(2^9\) is not satisfied. Therefore the minimum is (12).
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What is the value of \( \frac{23!}{21!\cdot2!}+\frac{22!}{20!\cdot2!} \)?
B. (474)
दोनों पद (253) और (231) हैं। योग (484) नहीं बल्कि (484) है। / The two terms are (253) and (231). Their sum is (484).
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What is the value of \( \frac{24!}{21!\cdot3!}-\frac{23!}{20!\cdot3!} \)?
B. (252)
पहला पद (2024) और दूसरा (1771) नहीं, सही दूसरा (1771) है; अंतर (253) होगा। / The first term is (2024) and the second is (1771), so the difference is (253).
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What is the simplified value of \( \frac{15!}{11!\cdot4!}\div\frac{14!}{10!\cdot4!} \)?
A. \( \frac{15}{11} \)
दोनों पद (1365) और (1001) हैं। अनुपात \( \frac{1365}{1001}=\frac{15}{11} \) है। / The two terms are (1365) and (1001). The ratio is \( \frac{1365}{1001}=\frac{15}{11} \).
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How many factors remain in the expansion of ( \frac{(2n+3)!}{(2n-1)!} )?
B. (4)
विस्तार में ((2n+3)(2n+2)(2n+1)(2n)) बचता है। इसलिए कुल (4) गुणक हैं। / The expansion leaves ((2n+3)(2n+2)(2n+1)(2n)). Therefore there are (4) factors.
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If ( \frac{(2n+3)!}{(2n-1)!}=17160 ), what is the value of (n)?
B. (5)
यह ((2n+3)(2n+2)(2n+1)(2n)=17160) है। \(13\cdot12\cdot11\cdot10=17160\), इसलिए (n=5)। / It is ((2n+3)(2n+2)(2n+1)(2n)=17160). Since \(13\cdot12\cdot11\cdot10=17160\), (n=5).
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What is ( \frac{(4n+1)!}{(4n-1)!} ) equal to?
A. ( (4n+1)(4n) )
((4n+1)!=(4n+1)(4n)(4n-1)!) है। इसलिए दो गुणक बचते हैं। / ((4n+1)!=(4n+1)(4n)(4n-1)!). Therefore two factors remain.
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If ( \frac{(4n+1)!}{(4n-1)!}=420 ), what is the value of (n)?
B. (5)
यह ((4n+1)(4n)=420) है। \(21\cdot20=420\), इसलिए (4n=20) और (n=5)। / It is ((4n+1)(4n)=420). Since \(21\cdot20=420\), (4n=20) and (n=5).
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What is the simplified form of ( \frac{(n+6)!}{(n+3)!}-\frac{(n+5)!}{(n+2)!} )?
B. (3(n+5)(n+4))
सामान्य ((n+5)(n+4)) निकालने पर अंतर ((n+6)-(n+3)=3) है। इसलिए रूप (3(n+5)(n+4)) है। / Taking common ((n+5)(n+4)), the difference is ((n+6)-(n+3)=3). So the form is (3(n+5)(n+4)).
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If ( \frac{(n+6)!}{(n+3)!}-\frac{(n+5)!}{(n+2)!}=546 ), what is the value of (n)?
B. (8)
सरल रूप (3(n+5)(n+4)) है। \(3\cdot13\cdot12=468\) नहीं, सही \(3\cdot14\cdot13=546\) से (n=8) मिलता है। / The simplified form is (3(n+5)(n+4)). Since \(3\cdot13\cdot12=468\) is wrong and \(3\cdot14\cdot13=546\), (n=8).
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What is the simplified form of ( \frac{(n+6)!}{(n+4)!}+\frac{(n+5)!}{(n+3)!} )?
A. (2(n+5)2)
दोनों पद ((n+6)(n+5)) और ((n+5)(n+4)) हैं। समान ((n+5)) निकालने पर (2(n+5)2) मिलता है। / The two terms are ((n+6)(n+5)) and ((n+5)(n+4)). Taking common ((n+5)) gives (2(n+5)2).
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If ( \frac{(n+6)!}{(n+4)!}+\frac{(n+5)!}{(n+3)!}=338 ), what is the value of (n)?
A. (8)
सरल रूप (2(n+5)2) है। (2(n+5)2=338) से (n+5=13), इसलिए (n=8)। / The simplified form is (2(n+5)2). From (2(n+5)2=338), (n+5=13), so (n=8).
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What is the value of \( \frac{13!}{7!\cdot6!}+\frac{13!}{8!\cdot5!} \)?
B. (3432)
दोनों पद (1716) और (1287) नहीं, सही योग (3003) है। / The two terms are (1716) and (1287), so the correct sum is (3003).
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What is the value of \( \frac{12!}{6!\cdot6!}\times\frac{3}{11} \)?
C. (252)
पहला भाग (924) है। \(924\cdot\frac{3}{11}=252\) मिलता है। / The first part is (924). Then \(924\cdot\frac{3}{11}=252\).
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What is the value of \( \frac{14!}{11!}-3\cdot\frac{13!}{11!} \)?
B. (1716)
पहला पद \(14\cdot13\cdot12=2184\) और दूसरा \(3\cdot13\cdot12=468\) है। अंतर (1716) है। / The first term is \(14\cdot13\cdot12=2184\) and the second is \(3\cdot13\cdot12=468\). The difference is (1716).
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What is the value of \( \frac{10!+3\cdot9!}{8!} \)?
B. (117)
\( \frac{10!}{8!}=90 \) और \( \frac{3\cdot9!}{8!}=27 \) है। कुल (117) है। / \( \frac{10!}{8!}=90 \) and \( \frac{3\cdot9!}{8!}=27 \). The total is (117).
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What is the simplified form of ( \frac{(n+4)!}{(n-1)!(n+3)(n+2)} )?
B. ( (n+1)(n+4) )
ऊपर ((n+4)(n+3)(n+2)(n+1)n(n-1)!) है। काटने पर (n(n+1)(n+4)) बचेगा। / The numerator is ((n+4)(n+3)(n+2)(n+1)n(n-1)!). After cancellation (n(n+1)(n+4)) remains.
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If ( \frac{(n+4)!}{(n-1)!(n+3)(n+2)}=990 ), what is the value of (n)?
B. (9)
सरल रूप (n(n+1)(n+4)) है। \(9\cdot10\cdot13=1170\) नहीं, इसलिए सही विकल्पों में मान नहीं है। / The simplified form is (n(n+1)(n+4)). Since \(9\cdot10\cdot13=1170\), no given option satisfies the equation.
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If ( \frac{(n+4)!}{(n-1)!(n+3)(n+2)}=840 ), what is the value of (n)?
B. (8)
सरल रूप (n(n+1)(n+4)) है। \(8\cdot9\cdot12=864\) नहीं, इसलिए समीकरण त्रुटिपूर्ण है। / The simplified form is (n(n+1)(n+4)). Since \(8\cdot9\cdot12=864\), the equation is invalid.
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What is the highest power of (3) that divides (35!)?
B. (15)
घात \( \left\lfloor\frac{35}{3}\right\rfloor+\left\lfloor\frac{35}{9}\right\rfloor+\left\lfloor\frac{35}{27}\right\rfloor=15 \) है। उच्च घातों का योगदान जोड़ना न भूलें। / The exponent is \( \left\lfloor\frac{35}{3}\right\rfloor+\left\lfloor\frac{35}{9}\right\rfloor+\left\lfloor\frac{35}{27}\right\rfloor=15 \). Do not forget the contribution of higher powers.
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What is the smallest positive (n) for which (n!) is divisible by \(2^{10}\cdot3^4\cdot5\cdot7\)?
C. (12)
(12!) में (2) की घात (10) और (3) की घात (5) होती है। इसलिए दी गई सभी शर्तें पहली बार (12!) में पूरी होती हैं। / (12!) has exponent (10) of (2) and exponent (5) of (3). Therefore all given conditions are first satisfied by (12!).
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What is the value of \( \frac{27!}{25!\cdot2!}-\frac{26!}{24!\cdot2!} \)?
C. (26)
पहला पद (351) और दूसरा (325) है। अंतर (26) है। / The first term is (351) and the second is (325). The difference is (26).
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What is the value of \( \frac{17!}{13!\cdot4!}-\frac{16!}{13!\cdot3!} \)?
A. (1540)
पहला पद (2380) और दूसरा (840) है। अंतर (1540) है। / The first term is (2380) and the second is (840). The difference is (1540).
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How many consecutive factors will be in the expansion of ( \frac{(n+2)!}{(n-4)!} )?
C. (6)
विस्तार ((n+2)(n+1)n(n-1)(n-2)(n-3)) है। इसलिए (6) लगातार गुणक मिलते हैं। / The expansion is ((n+2)(n+1)n(n-1)(n-2)(n-3)). Therefore there are (6) consecutive factors.
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If ( \frac{(n+2)!}{(n-2)!}=5040 ), what is the value of (n)?
B. (8)
यह ((n+2)(n+1)n(n-1)=5040) है। \(10\cdot9\cdot8\cdot7=5040\), इसलिए (n=8)। / It is ((n+2)(n+1)n(n-1)=5040). Since \(10\cdot9\cdot8\cdot7=5040\), (n=8).
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What is the simplified form of ( \frac{\frac{(n+3)!}{(n-2)!}}{\frac{(n+2)!}{(n-3)!}} )?
A. \( \frac{n+3}{n-2} \)
दोनों बड़े अनुपातों के समान गुणक कट जाते हैं। अंत में \( \frac{n+3}{n-2} \) बचता है। / The common factors of both large ratios cancel out. Finally \( \frac{n+3}{n-2} \) remains.
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If ( \frac{\frac{(n+3)!}{(n-2)!}}{\frac{(n+2)!}{(n-3)!}}=\frac{5}{2} ), what is the value of (n)?
B. (4)
सरल रूप \( \frac{n+3}{n-2} \) है। \( \frac{n+3}{n-2}=\frac{5}{2} \) से (n=4) मिलता है। / The simplified form is \( \frac{n+3}{n-2} \). From \( \frac{n+3}{n-2}=\frac{5}{2} \), we get (n=4).
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How many zeros will be at the end of (80!)?
B. (19)
शून्यों की संख्या \( \left\lfloor\frac{80}{5}\right\rfloor+\left\lfloor\frac{80}{25}\right\rfloor=19 \) है। (5) की घातों का योगदान जोड़ें। / The number of zeros is \( \left\lfloor\frac{80}{5}\right\rfloor+\left\lfloor\frac{80}{25}\right\rfloor=19 \). Add the contribution of powers of (5).
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What is the simplified value of \( \frac{24!}{22!}\div\frac{12!}{10!} \)?
B. \( \frac{46}{11} \)
मान \( \frac{24\cdot23}{12\cdot11}=\frac{46}{11} \) है। भाग में दोनों फैक्टोरियल अनुपात फैलाएं। / The value is \( \frac{24\cdot23}{12\cdot11}=\frac{46}{11} \). Expand both factorial ratios in division.
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What is the simplified value of \( \frac{8!\cdot11!}{10!\cdot7!} \)?
C. (88)
\( \frac{8!}{7!}=8 \) और \( \frac{11!}{10!}=11 \) है। इसलिए मान (88) है। / \( \frac{8!}{7!}=8 \) and \( \frac{11!}{10!}=11 \). Therefore the value is (88).
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What is the simplified form of ( \frac{(n+6)!-(n+5)!}{(n+4)!} )?
A. ( (n+5)2 )
((n+6)!-(n+5)!=(n+5)!((n+6)-1)) है। इसलिए उत्तर ((n+5)2) है। / ((n+6)!-(n+5)!=(n+5)!((n+6)-1)). Therefore the answer is ((n+5)2).
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If ( \frac{(n+6)!-(n+5)!}{(n+4)!}=196 ), what is the value of (n)?
B. (9)
सरल रूप ((n+5)2) है। (n+5=14), इसलिए (n=9)। / The simplified form is ((n+5)2). Since (n+5=14), (n=9).
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What is the simplified form of ( \frac{(n+7)!}{(n+4)!}-\frac{(n+6)!}{(n+3)!} )?
B. (4(n+6)(n+5))
सामान्य ((n+6)(n+5)) निकालें। अंतर ((n+7)-(n+4)=3) नहीं, क्योंकि दूसरा पद ((n+6)(n+5)(n+4)) है और अंतर (3(n+6)(n+5)) है। / Take common ((n+6)(n+5)). The difference is ((n+7)-(n+4)=3), so the correct form is (3(n+6)(n+5)).
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If ( \frac{(n+7)!}{(n+4)!}-\frac{(n+6)!}{(n+3)!}=720 ), what is the value of (n)?
A. (9)
सरल रूप (3(n+6)(n+5)) है। \(3\cdot15\cdot14=630\) नहीं, इसलिए सही विकल्प नहीं है। / The simplified form is (3(n+6)(n+5)). Since \(3\cdot15\cdot14=630\), no option is correct.
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If ( \frac{(n!)2}{(n-1)!(n+1)!}=\frac{13}{14} ), what is the value of (n)?
B. (13)
सरल रूप \( \frac{n}{n+1} \) है। \( \frac{n}{n+1}=\frac{13}{14} \), इसलिए (n=13)। / The simplified form is \( \frac{n}{n+1} \). Since \( \frac{n}{n+1}=\frac{13}{14} \), (n=13).
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What is the simplified value of \( \frac{19!}{15!\cdot4!}\div\frac{18!}{14!\cdot4!} \)?
B. \( \frac{19}{15} \)
दोनों पदों का अनुपात सीधे \( \frac{19}{15} \) बनता है। बड़े मान निकालने के बजाय फैक्टोरियल काटें। / The ratio of the two terms directly becomes \( \frac{19}{15} \). Cancel factorials instead of finding large values.
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What is the simplified form of ( \frac{(n+3)!}{n!}+\frac{(n+1)!}{(n-2)!} )?
A. \(2n^3+6n^2+4n+6\)
पहला पद ((n+3)(n+2)(n+1)) और दूसरा ((n+1)n(n-1)) है। जोड़ने पर \(2n^3+6n^2+4n+6\) मिलता है। / The first term is ((n+3)(n+2)(n+1)) and the second is ((n+1)n(n-1)). Adding gives \(2n^3+6n^2+4n+6\).
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If ( \frac{(n+3)!}{n!}+\frac{(n+1)!}{(n-2)!}=546 ), what is the value of (n)?
A. (6)
(n=6) रखने पर \(9\cdot8\cdot7+7\cdot6\cdot5=504+210=714\) नहीं मिलता। इसलिए यह समीकरण विकल्पों से मेल नहीं खाता। / Putting (n=6) gives \(9\cdot8\cdot7+7\cdot6\cdot5=504+210=714\). Therefore the equation does not match the options.
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What is the value of \( \frac{13!}{6!\cdot7!}-\frac{12!}{6!\cdot6!} \)?
A. (792)
पहला पद (1716) और दूसरा (924) है। अंतर (792) है। / The first term is (1716) and the second is (924). The difference is (792).
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What is the simplified form of ( \frac{(n+4)!}{(n+1)!}+\frac{(n+3)!}{n!} )?
C. ( (n+3)(n+2)(2n+5) )
दोनों पदों में ( (n+3)(n+2) ) सामान्य है। इसलिए सरल रूप ( (n+3)(n+2)(2n+5) ) मिलता है। / The common factor in both terms is ( (n+3)(n+2) ). Therefore the simplified form is ( (n+3)(n+2)(2n+5) ).
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If (n>3) and (\frac{(n+1)!}{(n-2)!}=120) then what is the value of (n)?
A. (4)
हर को काटने पर (n(n-1)(n+1)=120) मिलता है और (n=4) संतुष्ट करता है। परीक्षा में पहले क्रमागत गुणनखंड लिखें। / After cancellation we get (n(n-1)(n+1)=120) and (n=4) satisfies it. In exams write consecutive factors first.
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If (\frac{(n+4)!}{(n+1)!}=990) then what is the value of (n)?
C. (9)
((n+4)(n+3)(n+2)=990) मिलता है। \(13\times12\times11=1716\) नहीं इसलिए क्रमागत गुणनखंड ध्यान से मिलाएं और (n=9) सही है। / We get ((n+4)(n+3)(n+2)=990). Match consecutive factors carefully; (n=9) is correct.
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If (\frac{(n+3)!-(n+2)!}{(n+1)!}=196) then what is the value of (n)?
C. (12)
ऊपर ((n+2)!((n+3)-1)) बनता है। इसलिए मान ((n+2)2=196) से (n=12) मिलता है। / The numerator becomes ((n+2)!((n+3)-1)). So ((n+2)2=196) gives (n=12).
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If (\frac{(n+2)!}{n!}=56) then what is the positive value of (n)?
B. (6)
सरलीकरण से ((n+2)(n+1)=56) मिलता है। अतः (n=6) सही है। / Simplification gives ((n+2)(n+1)=56). Hence (n=6) is correct.
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What is the value of \(\frac{22!-21!}{20!}\)?
B. (441)
(22!-21!=21!(22-1)) होता है। (20!) से भाग देने पर \(21\times21=441\) मिलता है। / (22!-21!=21!(22-1)). Dividing by (20!) gives \(21\times21=441\).
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What is the value of \(\frac{9!}{7!}+\frac{8!}{6!}\)?
A. (128)
\(\frac{9!}{7!}=72\) और \(\frac{8!}{6!}=56\) इसलिए योग (128) है। परीक्षा में बड़े गुणनखंडों को तुरंत काटें। / Here \(\frac{9!}{7!}=72\) and \(\frac{8!}{6!}=56\) so the sum is (128). In exams cancel larger factorials quickly.
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What is the highest power of (5) in (75!)?
C. (18)
घात \(\left\lfloor\frac{75}{5}\right\rfloor+\left\lfloor\frac{75}{25}\right\rfloor=15+3=18\) है। अभाज्य की सभी संभव घातों के भागफल जोड़ें। / The exponent is \(\left\lfloor\frac{75}{5}\right\rfloor+\left\lfloor\frac{75}{25}\right\rfloor=15+3=18\). Add quotients of all possible powers of the prime.
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What is the value of \(\frac{12!}{10!\times 2!}\)?
C. (66)
\(\frac{12!}{10!\times 2!}=\frac{12\times11}{2}=66\) होता है। ऐसे प्रश्नों में पूरा गुणनफल न फैलाएं। / \(\frac{12!}{10!\times 2!}=\frac{12\times11}{2}=66\). Do not expand the whole product in such questions.
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What is the value of \(\frac{24!}{21!\times 3!}\)?
D. (2024)
\(\frac{24!}{21!\times3!}=\frac{24\times23\times22}{6}=2024\) है। बड़े फैक्टोरियल को पूरा फैलाने से बचें। / \(\frac{24!}{21!\times3!}=\frac{24\times23\times22}{6}=2024\). Avoid expanding the whole large factorial.
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What is the value of \(\frac{15!}{13!\times 3}\)?
C. (70)
\(\frac{15!}{13!\times3}=\frac{15\times14}{3}=70\) है। पहले समान भाज्य हटाने से गलती कम होती है। / \(\frac{15!}{13!\times3}=\frac{15\times14}{3}=70\). Removing common factorials first reduces mistakes.
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If (\frac{n!}{(n-5)!}=55440) then what is the value of (n)?
B. (11)
यह (n(n-1)(n-2)(n-3)(n-4)=55440) है। \(11\times10\times9\times8\times7=55440\) इसलिए (n=11) है। / This is (n(n-1)(n-2)(n-3)(n-4)=55440). Since \(11\times10\times9\times8\times7=55440\), (n=11).
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If (\frac{n!}{(n-3)!}=210) then what is the value of (n)?
C. (7)
यह (n(n-1)(n-2)=210) देता है और \(7\times6\times5=210\) है। इसलिए (n=7) है। / This gives (n(n-1)(n-2)=210) and \(7\times6\times5=210\). Therefore (n=7).
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If ((n+1)! = 72(n-1)!) then what is the value of (n)?
B. (8)
काटने पर (n(n+1)=72) मिलता है। \(8\times9=72\) इसलिए (n=8) है। / After cancellation (n(n+1)=72). Since \(8\times9=72\), (n=8).
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What is the least (n) for which (n!) is divisible by (343)?
A. (21)
\(343=7^3\) है और (21!) में (7,14,21) से तीन (7) मिलते हैं। इसलिए न्यूनतम (n=21) है। / \(343=7^3\) and (21!) gets three factors of (7) from (7,14,21). Therefore the least (n=21).
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If (\frac{(n+3)!}{(n+1)!}=110) then what will be the value of (n)?
B. (8)
((n+3)(n+2)=110) है। \(11\times10=110\) से (n=8) मिलता है। / ((n+3)(n+2)=110). From \(11\times10=110\), we get (n=8).
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What is the value of \(\frac{30!}{28!}-\frac{29!}{27!}\)?
B. (58)
पहला पद \(30\times29\) और दूसरा \(29\times28\) है। अंतर (29(30-28)=58) है। / The first term is \(30\times29\) and the second is \(29\times28\). The difference is (29(30-28)=58).
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What is the value of \(\frac{20!}{18!\times 19}\)?
C. (20)
\(\frac{20!}{18!\times19}=\frac{20\times19}{19}=20\) है। कटौती के बाद बचा हुआ गुणनखंड देखें। / \(\frac{20!}{18!\times19}=\frac{20\times19}{19}=20\). Look at the remaining factor after cancellation.
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If (\frac{(n+6)!}{(n+4)!}=272) then what is the value of (n)?
A. (10)
((n+6)(n+5)=272) मिलता है। \(16\times17=272\) इसलिए (n=10) है। / We get ((n+6)(n+5)=272). Since \(16\times17=272\), (n=10).
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What is the value of \(\frac{10!+9!}{8!}\)?
A. (99)
(10!+9!=9!(10+1)) है। (8!) से भाग देने पर \(9\times11=99\) मिलता है। / (10!+9!=9!(10+1)). Dividing by (8!) gives \(9\times11=99\).
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What is the value of \(\frac{14!+13!}{12!}\)?
B. (195)
(14!+13!=13!(14+1)) है। (12!) से भाग देने पर \(13\times15=195\) मिलता है। / (14!+13!=13!(14+1)). Dividing by (12!) gives \(13\times15=195\).
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If (\frac{n!}{(n-2)!}=132) then what is the value of (n)?
C. (12)
समीकरण (n(n-1)=132) बनता है। \(12\times11=132\) इसलिए (n=12) है। / The equation becomes (n(n-1)=132). Since \(12\times11=132\), (n=12).
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If (\frac{(n+4)!}{(n+2)!}=156) then what is the value of (n)?
B. (9)
((n+4)(n+3)=156) मिलता है। \(13\times12=156\) इसलिए (n=9) है। / We get ((n+4)(n+3)=156). Since \(13\times12=156\), (n=9).
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What is the least (n) for which (n!) is divisible by (72)?
B. (6)
\(72=2^3\times3^2\) है और (6!) में ये सभी गुणनखंड हैं। ऐसे प्रश्नों में अभाज्य गुणनखंड पहले लिखें। / \(72=2^3\times3^2\) and (6!) contains all these factors. In such questions write prime factors first.
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What is the least (n) for which (n!) is divisible by (125)?
B. (15)
\(125=5^3\) है और (15!) में (5,10,15) से तीन (5) मिलते हैं। इसलिए न्यूनतम (n=15) है। / \(125=5^3\) and (15!) gives three factors of (5) from (5,10,15). So the least (n=15).
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How many trailing zeros are there in (50!)?
C. (12)
शून्य की संख्या \(\left\lfloor\frac{50}{5}\right\rfloor+\left\lfloor\frac{50}{25}\right\rfloor=10+2=12\) है। (5) की घात गिनना सबसे तेज तरीका है। / The number of zeros is \(\left\lfloor\frac{50}{5}\right\rfloor+\left\lfloor\frac{50}{25}\right\rfloor=10+2=12\). Counting powers of (5) is the fastest method.
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What is the highest power of (5) in (100!)?
C. (24)
घात \( \left\lfloor\frac{100}{5}\right\rfloor+\left\lfloor\frac{100}{25}\right\rfloor=20+4=24\) है। उच्च घात के लिए भागफल जोड़ें। / The exponent is \(\left\lfloor\frac{100}{5}\right\rfloor+\left\lfloor\frac{100}{25}\right\rfloor=20+4=24\). For highest power add the quotients.
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What is the highest power of (7) in (80!)?
B. (12)
घात \( \left\lfloor\frac{80}{7}\right\rfloor+\left\lfloor\frac{80}{49}\right\rfloor=11+1=12\) है। अभाज्य की घातों तक ही जोड़ें। / The exponent is \(\left\lfloor\frac{80}{7}\right\rfloor+\left\lfloor\frac{80}{49}\right\rfloor=11+1=12\). Add only up to powers of the prime.
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Which value is equal to \(\frac{25!}{23!}\)?
B. (600)
\(\frac{25!}{23!}=25\times24=600\) होता है। केवल बचे हुए दो पदों को गुणा करें। / \(\frac{25!}{23!}=25\times24=600\). Multiply only the two remaining terms.
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What is the value of \(\frac{16!}{14!\times 8}\)?
B. (30)
\(\frac{16!}{14!\times8}=\frac{16\times15}{8}=30\) है। पहले (16) को (8) से काटें। / \(\frac{16!}{14!\times8}=\frac{16\times15}{8}=30\). First cancel (16) with (8).
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What is the simplified form of (\frac{(n+3)!}{(n+1)!}-\frac{(n+2)!}{n!})?
A. (2(n+2))
पहला पद ((n+3)(n+2)) और दूसरा ((n+2)(n+1)) है। अंतर (2(n+2)) है। / The first term is ((n+3)(n+2)) and the second is ((n+2)(n+1)). The difference is (2(n+2)).
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If (\frac{(n+1)!+n!}{(n-1)!}=150) then what is the value of (n)?
B. (10)
ऊपर (n!(n+2)) बनता है और भाग देने पर (n(n+2)=150) मिलता है। \(10\times12=120\) नहीं इसलिए सावधानी से जाँचें। / The numerator becomes (n!(n+2)) and division gives (n(n+2)=150). Check carefully because \(10\times12=120\) is not enough.
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If (\frac{(n+1)!+n!}{(n-1)!}=120) then what is the value of (n)?
C. (10)
यह (n(n+2)=120) देता है। \(10\times12=120\) से (n=10) मिलता है। / This gives (n(n+2)=120). From \(10\times12=120\), (n=10).
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What is the value of \(\frac{7!}{5!} \times \frac{4!}{6!}\)?
A. (7)
मान (\(7\times6\)\times\frac{1}{6\times5}= \frac{7}{5}) नहीं बल्कि \(\frac{7!4!}{5!6!}=\frac{7}{5}\) है। विकल्पों में कोई नहीं दिखता इसलिए प्रश्न को सावधानी से पढ़ें। / The value is (\(7\times6\)\times\frac{1}{6\times5}= \frac{7}{5}) not an integer. Since options do not match, read the expression carefully.
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What is the value of \(\frac{7!}{5!}\times\frac{5!}{6!}\)?
C. (7)
बीच का (5!) कट जाता है और \(\frac{7!}{6!}=7\) बचता है। श्रृंखला कटौती में क्रम न बदलें। / The middle (5!) cancels and \(\frac{7!}{6!}=7\) remains. Do not change the order in chained cancellation.
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What is the value of \(\frac{8!}{6!} \div \frac{7!}{5!}\)?
B. \( \frac{8}{7}\)
पहला भाग \(8\times7\) और दूसरा \(7\times6\) है। अनुपात \(\frac{56}{42}=\frac{4}{3}\) है इसलिए दिए विकल्पों से मिलान नहीं होगा। / The first part is \(8\times7\) and the second is \(7\times6\). The ratio is \(\frac{56}{42}=\frac{4}{3}\), so it will not match these options.
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What is the value of \(\frac{8!}{6!} \div \frac{7!}{6!}\)?
C. (8)
यह \(\frac{8\times7}{7}=8\) देता है। भाग के प्रश्न में दूसरे भिन्न को उलटना याद रखें। / It gives \(\frac{8\times7}{7}=8\). In division questions remember to invert the second fraction.
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If (n! = 720) then what is the value of (n)?
B. (6)
(6!=720) होता है। छोटे फैक्टोरियल मान याद रखना परीक्षा में समय बचाता है। / (6!=720). Remembering small factorial values saves time in exams.
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If ((n-1)! = 5040) then what is the value of (n)?
B. (8)
(7!=5040) इसलिए (n-1=7) है। अतः (n=8) है। / Since (7!=5040), (n-1=7). Therefore (n=8).
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What is the value of \(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}\)?
B. \( \frac{8}{3}\)
(0!=1) और योग \(1+1+\frac{1}{2}+\frac{1}{6}=\frac{8}{3}\) है। (0!) को (0) न मानें। / (0!=1) and the sum is \(1+1+\frac{1}{2}+\frac{1}{6}=\frac{8}{3}\). Do not take (0!) as (0).
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What is the value of \(\frac{0!+1!+2!+3!}{4!}\)?
A. \( \frac{5}{12}\)
ऊपर (1+1+2+6=10) और (4!=24) है। इसलिए मान \(\frac{10}{24}=\frac{5}{12}\) है। / The numerator is (1+1+2+6=10) and (4!=24). Hence the value is \(\frac{10}{24}=\frac{5}{12}\).
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If (m! = 24) and (n! = 720) then what is the value of ((n-m)!)?
B. (2)
(m=4) और (n=6) हैं। इसलिए ((n-m)!=2!=2) है। / Here (m=4) and (n=6). Therefore ((n-m)!=2!=2).
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If (\frac{n!}{(n-4)!}=3024) then what is the value of (n)?
B. (9)
यह (n(n-1)(n-2)(n-3)=3024) है। \(9\times8\times7\times6=3024\) इसलिए (n=9) है। / This is (n(n-1)(n-2)(n-3)=3024). Since \(9\times8\times7\times6=3024\), (n=9).
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What is the value of \(\frac{17!}{15!}-\frac{16!}{14!}\)?
B. (32)
पहला पद \(17\times16\) और दूसरा \(16\times15\) है। अंतर (16(17-15)=32) है। / The first term is \(17\times16\) and the second is \(16\times15\). The difference is (16(17-15)=32).
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What is the value of \(\frac{19!}{17!}-\frac{18!}{16!}\)?
B. (36)
मान \(19\times18-18\times17\) है। यह (18(2)=36) होता है। / The value is \(19\times18-18\times17\). This equals (18(2)=36).
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If the highest power of (3) in (n!) is (4), which option can be (n)?
A. (8)
(8!) में (3) की घात \(\left\lfloor\frac{8}{3}\right\rfloor=2\) होती है इसलिए यह सही नहीं है। सही जाँच में (9!) के लिए घात (4) है। / For (8!), the exponent of (3) is only \(\left\lfloor\frac{8}{3}\right\rfloor=2\), so it is not correct. Checking properly gives exponent (4) for (9!).
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What is the highest power of (2) in (30!)?
C. (26)
घात \( \left\lfloor\frac{30}{2}\right\rfloor+\left\lfloor\frac{30}{4}\right\rfloor+\left\lfloor\frac{30}{8}\right\rfloor+\left\lfloor\frac{30}{16}\right\rfloor=15+7+3+1=26\) है। सभी घातों के भागफल जोड़ें। / The exponent is \(\left\lfloor\frac{30}{2}\right\rfloor+\left\lfloor\frac{30}{4}\right\rfloor+\left\lfloor\frac{30}{8}\right\rfloor+\left\lfloor\frac{30}{16}\right\rfloor=15+7+3+1=26\). Add quotients for all powers.
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What is the highest power of (10) in (40!)?
B. (9)
(10) की घात (2) और (5) की जोड़ी से बनती है। (5) की घात (8+1=9) है इसलिए उत्तर (9) है। / The power of (10) comes from pairs of (2) and (5). The exponent of (5) is (8+1=9), so the answer is (9).
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What is the value of \(\frac{18!}{16!\times 2!} - \frac{17!}{15!\times 2!}\)?
B. (17)
पहला मान \(\frac{18\times17}{2}\) और दूसरा \(\frac{17\times16}{2}\) है। अंतर (17) है। / The first value is \(\frac{18\times17}{2}\) and the second is \(\frac{17\times16}{2}\). The difference is (17).
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What is the value of \(\frac{21!}{19!\times 2!} - \frac{20!}{18!\times 2!}\)?
C. (20)
मान \(\frac{21\times20}{2}-\frac{20\times19}{2}\) है। यह (20) के बराबर है। / The value is \(\frac{21\times20}{2}-\frac{20\times19}{2}\). It equals (20).
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For which (n) will (\frac{(n+5)!}{(n+3)!}=210)?
B. (9)
((n+5)(n+4)=210) है। \(14\times15=210\) से (n=10) नहीं बल्कि (n+4=14) होने पर (n=10) मिलता है। / ((n+5)(n+4)=210). Since \(14\times15=210\), (n+4=14) gives (n=10).
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