permutations se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
967 questions tagged with permutations.
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+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+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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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 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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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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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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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{(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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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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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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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 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 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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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 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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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 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 \( \frac{10!}{5!\cdot5!}\times\frac{5}{9} \)?
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 \( \frac{12!}{8!\cdot4!}+\frac{12!}{9!\cdot3!} \)?
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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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 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+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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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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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{(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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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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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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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 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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