Level 56 • 50/50 questions • 30 seconds per question.
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What is the value of \( \frac{12!}{9!} \)?
C. (1320)
(12!) को \(12\cdot11\cdot10\cdot9!\) लिखने पर मान (1320) मिलता है। परीक्षा में बड़े फैक्टोरियल को छोटे फैक्टोरियल तक फैलाएं। / Writing (12!) as \(12\cdot11\cdot10\cdot9!\) gives (1320). In exams, expand the larger factorial up to the smaller factorial.
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What is the simplified value of \( \frac{10!}{7!\cdot3!} \)?
B. (120)
(10!) को \(10\cdot9\cdot8\cdot7!\) लिखकर (3!) से भाग देने पर (120) मिलता है। समान फैक्टोरियल पहले काटें। / Writing (10!) as \(10\cdot9\cdot8\cdot7!\) and dividing by (3!) gives (120). Cancel common factorials first.
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Find the value of \( \frac{8!-7!}{6!} \).
D. (49)
(8!-7!=7!(8-1)) और \( \frac{7!}{6!}=7 \), इसलिए मान (49) है। घटाव में सामान्य फैक्टर निकालें। / (8!-7!=7!(8-1)) and \( \frac{7!}{6!}=7 \), so the value is (49). Take the common factor in subtraction.
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If ( \frac{(n+2)!}{n!}=90 ), what is the value of (n)?
A. (8)
यह ((n+2)(n+1)=90) देता है और \(10\cdot9=90\), इसलिए (n=8)। लगातार गुणकों को पहचानना तेज तरीका है। / It gives ((n+2)(n+1)=90), and \(10\cdot9=90\), so (n=8). Recognizing consecutive factors is a quick method.
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If ( \frac{n!}{(n-3)!}=336 ), what is the positive value of (n)?
C. (8)
यह (n(n-1)(n-2)=336) है और \(8\cdot7\cdot6=336\)। ऐसे प्रश्नों में तीन लगातार घटते गुणक देखें। / It is (n(n-1)(n-2)=336), and \(8\cdot7\cdot6=336\). In such questions, look for three consecutive decreasing factors.
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What is the simplified form of ( \frac{(n+2)!+(n+1)!}{n!} )?
B. ( (n+1)(n+3) )
दोनों पदों में ((n+1)!) सामान्य है, इसलिए सरल रूप ((n+1)(n+3)) है। पहले सामान्य फैक्टोरियल बाहर निकालें। / ((n+1)!) is common in both terms, so the simplified form is ((n+1)(n+3)). Take the common factorial out first.
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If ( \frac{(n+3)!-(n+2)!}{(n+1)!}=49 ), what will be the value of (n)?
D. (5)
अभिव्यक्ति ((n+2)2) बनती है, इसलिए ((n+2)2=49) और (n=5)। पहले घटाव को सरल करें। / The expression becomes ((n+2)2), so ((n+2)2=49) and (n=5). Simplify the subtraction first.
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What is the simplified value of \( \frac{14!}{11!}\div\frac{7!}{4!} \)?
A. \( \frac{52}{5} \)
मान \( \frac{14\cdot13\cdot12}{7\cdot6\cdot5}=\frac{52}{5} \) है। भाग में दोनों फैक्टोरियल अनुपात अलग-अलग फैलाएं। / The value is \( \frac{14\cdot13\cdot12}{7\cdot6\cdot5}=\frac{52}{5} \). In division, expand both factorial ratios separately.
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What is the value of \( \frac{9!}{6!\cdot2!} \)?
C. (252)
(9!) को \(9\cdot8\cdot7\cdot6!\) लिखें और (2!) से भाग दें। इससे मान (252) आता है। / Write (9!) as \(9\cdot8\cdot7\cdot6!\) and divide by (2!). This gives (252).
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What is ( \frac{(2n+2)!}{(2n)!} ) equal to?
B. ( (2n+2)(2n+1) )
((2n+2)!=(2n+2)(2n+1)(2n)!), इसलिए उत्तर ((2n+2)(2n+1)) है। सूचक वाले फैक्टोरियल में क्रम न छोड़ें। / ((2n+2)!=(2n+2)(2n+1)(2n)!), so the answer is ((2n+2)(2n+1)). Do not skip the order in indexed factorials.
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If ( \frac{(2n+2)!}{(2n)!}=132 ), what is the value of (n)?
D. (5)
यह ((2n+2)(2n+1)=132) देता है और \(12\cdot11=132\), इसलिए (n=5)। पहले लगातार गुणकों को मिलाएं। / It gives ((2n+2)(2n+1)=132), and \(12\cdot11=132\), so (n=5). Match the consecutive factors first.
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What is the simplified form of ( \frac{m!}{(m-2)!(m-1)} )?
A. (m)
(m!=m(m-1)(m-2)!), इसलिए काटने पर (m) बचता है। फैक्टोरियल को ठीक उसी सीमा तक फैलाएं। / (m!=m(m-1)(m-2)!), so (m) remains after cancellation. Expand the factorial exactly up to the needed limit.
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If ( \frac{n!}{(n-2)!}=9n ), what is the positive value of (n)?
B. (10)
बायां पक्ष (n(n-1)) है, इसलिए (n(n-1)=9n)। (n>0) होने से (n-1=9) और (n=10)। / The left side is (n(n-1)), so (n(n-1)=9n). Since (n>0), (n-1=9) and (n=10).
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What is the value of \( \frac{11!}{8!\cdot3!}\div\frac{10!}{7!\cdot3!} \)?
C. \( \frac{11}{8} \)
पहला पद (165) और दूसरा (120) है, इसलिए अनुपात \( \frac{11}{8} \) है। भाग को भिन्न के अनुपात की तरह लें। / The first term is (165) and the second is (120), so the ratio is \( \frac{11}{8} \). Treat division as a ratio of fractions.
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What is the value of \( \frac{12!}{8!\cdot4!}-\frac{11!}{7!\cdot4!} \)?
A. (165)
दोनों पद क्रमशः (495) और (330) हैं, इसलिए अंतर (165) है। पहले हर पद को अलग सरल करें। / The two terms are (495) and (330), so the difference is (165). Simplify each term separately first.
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What is the simplified form of ( \frac{(n+1)!+n!}{n!} )?
D. (n+2)
((n+1)!=(n+1)n!) है, इसलिए योग ((n+2)n!) बनता है। भाग देने पर (n+2) मिलता है। / ((n+1)!=(n+1)n!), so the sum becomes ((n+2)n!). Dividing gives (n+2).
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If ( \frac{(n+1)!+n!}{n!}=13 ), what is the value of (n)?
C. (11)
सरल रूप (n+2) है, इसलिए (n+2=13) और (n=11)। पहले अभिव्यक्ति को छोटा करें। / The simplified form is (n+2), so (n+2=13) and (n=11). Reduce the expression first.
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What is the value of \( \frac{15!}{13!}-\frac{14!}{12!} \)?
B. (28)
पहला मान \(15\cdot14=210\) और दूसरा \(14\cdot13=182\) है। अंतर (28) है। / The first value is \(15\cdot14=210\) and the second is \(14\cdot13=182\). The difference is (28).
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What is the value of \( \frac{6!}{3!}+\frac{5!}{2!} \)?
D. (180)
दोनों पद (120) और (60) हैं, इसलिए योग (180) है। मिश्रित योग में हर पद अलग हल करें। / The two terms are (120) and (60), so the sum is (180). In a mixed sum, solve each term separately.
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If ( \frac{(n+4)!}{(n+2)!}+\frac{(n+3)!}{(n+1)!}=200 ), what is the value of (n)?
C. (7)
अभिव्यक्ति (2(n+3)2) बनती है, इसलिए (2(n+3)2=200)। इससे (n+3=10) और (n=7)। / The expression becomes (2(n+3)2), so (2(n+3)2=200). Thus (n+3=10) and (n=7).
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What is the simplified form of ( \frac{\frac{(n+3)!}{n!}}{n+3} )?
D. ( (n+2)(n+1) )
ऊपर का अनुपात ((n+3)(n+2)(n+1)) है। (n+3) से भाग देने पर ((n+2)(n+1)) बचता है। / The upper ratio is ((n+3)(n+2)(n+1)). Dividing by (n+3) leaves ((n+2)(n+1)).
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What is the value of \( \frac{13!}{10!\cdot3!}-\frac{12!}{9!\cdot3!} \)?
B. (66)
पहला पद (286) और दूसरा (220) है, इसलिए अंतर (66) है। ऐसे प्रश्नों में हर पद का मान लिखकर घटाएं। / The first term is (286) and the second is (220), so the difference is (66). In such questions, write each term value and subtract.
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What is the smallest positive (n) for which (n!) is divisible by (72)?
A. (6)
\(72=2^3\cdot3^2\) है और (6!) में ये सभी गुणक मिल जाते हैं। विभाज्यता में अभाज्य गुणनखंड जांचें। / \(72=2^3\cdot3^2\), and (6!) contains all these factors. For divisibility, check prime factors.
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How many zeros will be at the end of (18!)?
D. (3)
अंतिम शून्यों की संख्या (5) के गुणकों से मिलती है और (18!) में (5,10,15) हैं। इसलिए कुल (3) शून्य होंगे। / The number of ending zeros comes from multiples of (5), and (18!) has (5,10,15). Therefore, there are (3) zeros.
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What is the highest power of (2) that divides (10!)?
C. (8)
घात \( \left\lfloor\frac{10}{2}\right\rfloor+\left\lfloor\frac{10}{4}\right\rfloor+\left\lfloor\frac{10}{8}\right\rfloor=8 \) है। अभाज्य घात के लिए भागफल जोड़ें। / The exponent is \( \left\lfloor\frac{10}{2}\right\rfloor+\left\lfloor\frac{10}{4}\right\rfloor+\left\lfloor\frac{10}{8}\right\rfloor=8 \). For prime exponents, add the quotients.
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What will be the highest power of (5) in (25!)?
B. (6)
घात \( \left\lfloor\frac{25}{5}\right\rfloor+\left\lfloor\frac{25}{25}\right\rfloor=6 \) है। (25) जैसे उच्च गुणकों को न भूलें। / The exponent is \( \left\lfloor\frac{25}{5}\right\rfloor+\left\lfloor\frac{25}{25}\right\rfloor=6 \). Do not forget higher multiples such as (25).
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What is the simplified form of ( \frac{(n!)2}{(n-1)!(n+1)!} )?
C. \( \frac{n}{n+1} \)
( \frac{n!}{(n-1)!}=n ) और ( \frac{n!}{(n+1)!}=\frac{1}{n+1} ), इसलिए उत्तर \( \frac{n}{n+1} \) है। अनुपात को दो छोटे भागों में तोड़ें। / ( \frac{n!}{(n-1)!}=n ) and ( \frac{n!}{(n+1)!}=\frac{1}{n+1} ), so the answer is \( \frac{n}{n+1} \). Break the ratio into two smaller parts.
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If ( \frac{(n!)2}{(n-1)!(n+1)!}=\frac{5}{6} ), what is the value of (n)?
A. (5)
सरल रूप \( \frac{n}{n+1} \) है, इसलिए \( \frac{n}{n+1}=\frac{5}{6} \)। इससे (n=5) मिलता है। / The simplified form is \( \frac{n}{n+1} \), so \( \frac{n}{n+1}=\frac{5}{6} \). This gives (n=5).
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What is the simplified form of ( \frac{(n+2)!}{(n-2)!\cdot n(n-1)} )?
B. ( (n+2)(n+1) )
ऊपर ((n+2)(n+1)n(n-1)(n-2)!) बनता है। हर से काटने पर ((n+2)(n+1)) बचता है। / The numerator becomes ((n+2)(n+1)n(n-1)(n-2)!). After cancellation with the denominator, ((n+2)(n+1)) remains.
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If ( \frac{(n+2)!}{(n-2)!}=1680 ), what is the value of (n)?
D. (6)
यह ((n+2)(n+1)n(n-1)=1680) है। \(8\cdot7\cdot6\cdot5=1680\), इसलिए (n=6)। / It is ((n+2)(n+1)n(n-1)=1680). Since \(8\cdot7\cdot6\cdot5=1680\), (n=6).
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What is the value of \( \frac{16!}{14!\cdot2!}+\frac{15!}{13!\cdot2!} \)?
C. (225)
पहला पद (120) और दूसरा (105) है, इसलिए योग (225) है। जोड़ने से पहले दोनों भाग अलग निकालें। / The first term is (120) and the second is (105), so the sum is (225). Find both parts separately before adding.
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What is the simplified value of \( \frac{18!}{15!}\div\frac{6!}{3!} \)?
B. \( \frac{204}{5} \)
मान \( \frac{18\cdot17\cdot16}{6\cdot5\cdot4}=\frac{204}{5} \) है। भाग में दोनों पक्षों को समान लंबाई के गुणनफल में बदलें। / The value is \( \frac{18\cdot17\cdot16}{6\cdot5\cdot4}=\frac{204}{5} \). In division, convert both sides into products of equal length.
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What is ( \frac{(3n)!}{(3n-2)!} ) equal to?
A. (3n(3n-1))
((3n)!=(3n)(3n-1)(3n-2)!), इसलिए उत्तर (3n(3n-1)) है। सूचक वाले पद को एक इकाई की तरह संभालें। / ((3n)!=(3n)(3n-1)(3n-2)!), so the answer is (3n(3n-1)). Handle the indexed term as one unit.
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If ( \frac{(3n)!}{(3n-2)!}=132 ), what is the value of (n)?
D. (4)
यह (3n(3n-1)=132) है और \(12\cdot11=132\), इसलिए (3n=12)। अतः (n=4)। / It is (3n(3n-1)=132), and \(12\cdot11=132\), so (3n=12). Hence (n=4).
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Which is the correct expansion of ( \frac{(2n)!}{(2n-3)!} )?
C. ( (2n)(2n-1)(2n-2) )
((2n)!) को ((2n)(2n-1)(2n-2)(2n-3)!) तक फैलाते हैं। इसलिए तीन गुणक शेष रहते हैं। / We expand ((2n)!) up to ((2n)(2n-1)(2n-2)(2n-3)!). Therefore, three factors remain.
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If ( \frac{(2n)!}{(2n-3)!}=720 ), what is the value of (n)?
A. (5)
यह ((2n)(2n-1)(2n-2)=720) है। \(10\cdot9\cdot8=720\), इसलिए (2n=10) और (n=5)। / It is ((2n)(2n-1)(2n-2)=720). Since \(10\cdot9\cdot8=720\), (2n=10) and (n=5).
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By how much is \( \frac{9!}{7!} \) greater than \( \frac{8!}{6!} \)?
B. (16)
पहला मान (72) और दूसरा (56) है। अंतर (16) है। / The first value is (72) and the second is (56). The difference is (16).
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What is the simplified value of \( \frac{4!\cdot8!}{5!\cdot7!} \)?
D. \( \frac{8}{5} \)
\( \frac{4!}{5!}=\frac{1}{5} \) और \( \frac{8!}{7!}=8 \), इसलिए मान \( \frac{8}{5} \) है। गुणन में अलग-अलग अनुपात काटें। / \( \frac{4!}{5!}=\frac{1}{5} \) and \( \frac{8!}{7!}=8 \), so the value is \( \frac{8}{5} \). Cancel separate ratios in multiplication.
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What is the simplified form of ( \frac{(n+1)!}{(n-1)!}-\frac{n!}{(n-2)!} )?
C. (2n)
पहला पद (n(n+1)) और दूसरा (n(n-1)) है। अंतर (2n) है। / The first term is (n(n+1)) and the second is (n(n-1)). The difference is (2n).
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If ( \frac{(n+1)!}{(n-1)!}-\frac{n!}{(n-2)!}=18 ), what is the value of (n)?
B. (9)
सरल रूप (2n) है, इसलिए (2n=18)। अतः (n=9)। / The simplified form is (2n), so (2n=18). Hence (n=9).
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What is the simplified form of ( \frac{(n+2)!}{(n-1)!}-\frac{(n+1)!}{(n-2)!} )?
A. (3n(n+1))
पद (n(n+1)(n+2)) और (n(n+1)(n-1)) बनते हैं। अंतर (3n(n+1)) है। / The terms become (n(n+1)(n+2)) and (n(n+1)(n-1)). The difference is (3n(n+1)).
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If ( \frac{(n+2)!}{(n-1)!}-\frac{(n+1)!}{(n-2)!}=216 ), what is the value of (n)?
D. (8)
सरल रूप (3n(n+1)) है, इसलिए (n(n+1)=72)। \(8\cdot9=72\), इसलिए (n=8)। / The simplified form is (3n(n+1)), so (n(n+1)=72). Since \(8\cdot9=72\), (n=8).
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How many consecutive factors remain in the expansion of ( \frac{(n+5)!}{(n+1)!} )?
B. (4)
विस्तार ((n+5)(n+4)(n+3)(n+2)) है, इसलिए (4) गुणक बचते हैं। फैक्टोरियल अंतर से गुणकों की संख्या मिलती है। / The expansion is ((n+5)(n+4)(n+3)(n+2)), so (4) factors remain. The factorial gap gives the number of factors.
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When (n=3), what is the value of ( \frac{(n+5)!}{(n+1)!} )?
C. (1680)
(n=3) रखने पर \( \frac{8!}{4!}=8\cdot7\cdot6\cdot5=1680 \)। मान रखने से पहले सामान्य रूप समझें। / Putting (n=3) gives \( \frac{8!}{4!}=8\cdot7\cdot6\cdot5=1680 \). Understand the general form before substituting.
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What is the value of \( \frac{20!}{18!\cdot2!}-\frac{19!}{18!} \)?
A. (171)
पहला पद (190) और दूसरा (19) है। अंतर (171) है। / The first term is (190) and the second is (19). The difference is (171).
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What is the simplified value of \( \frac{11!}{5!\cdot6!}\div\frac{10!}{5!\cdot5!} \)?
D. \( \frac{11}{6} \)
दोनों पद (462) और (252) हैं, इसलिए अनुपात \( \frac{11}{6} \) है। बड़े मानों को काटकर सरल अनुपात बनाएं। / The two terms are (462) and (252), so the ratio is \( \frac{11}{6} \). Reduce large values to a simple ratio.
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If ( \frac{(n+4)!}{(n+2)!}=182 ), what is the value of (n)?
B. (10)
यह ((n+4)(n+3)=182) है और \(14\cdot13=182\)। इसलिए (n+4=14) और (n=10)। / It is ((n+4)(n+3)=182), and \(14\cdot13=182\). Therefore, (n+4=14) and (n=10).
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What is the value of \(\frac{13!}{10!}+\frac{12!}{9!}\)?
A. (3036)
पहला पद \(13\cdot12\cdot11=1716\) और दूसरा \(12\cdot11\cdot10=1320\) है। योग (3036) है। / The first term is \(13\cdot12\cdot11=1716\) and the second is \(12\cdot11\cdot10=1320\). The sum is (3036).
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If \( \frac{(n+5)!}{(n+3)!}=210\), what is the value of (n)?
B. (9)
यह \((n+5)(n+4)=210\) देता है और \(15\cdot14=210\)। इसलिए (n+5=15) और (n=10) नहीं, सही (n=10) है। / It gives \((n+5)(n+4)=210\), and \(15\cdot14=210\). Therefore, (n=10).
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What is the value of \(\frac{21!}{19!\cdot2!}-\frac{18!}{16!\cdot2!}\)?
C. (72)
पहला पद (210) और दूसरा (153) है। अंतर (57) है। / The first term is (210) and the second is (153). The difference is (57).
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