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 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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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 \(\frac{13!}{10!}+\frac{12!}{9!}\)?
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+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 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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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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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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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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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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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+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+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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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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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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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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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{(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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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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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 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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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 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-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-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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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 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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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 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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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 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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