Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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Showing 191-200 of 590 questions.
If (\frac{(n+1)!}{(n-1)!}=20), what is the value of (n)?
B. (4)
यह अनुपात (n(n+1)) है। \(4\times5=20\), इसलिए (n=4)। / This ratio is (n(n+1)). Since \(4\times5=20\), (n=4).
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What is the value of \(\frac{8!}{6!}-\frac{5!}{4!}+2!\)?
C. (53)
\(\frac{8!}{6!}=56\), \(\frac{5!}{4!}=5\) और (2!=2)। इसलिए (56-5+2=53) है। / \(\frac{8!}{6!}=56\), \(\frac{5!}{4!}=5\), and (2!=2). Thus (56-5+2=53).
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What is the value of \(\frac{7!}{4!,3!}+\frac{7!}{6!,1!}\)?
C. (42)
पहला पद (35) और दूसरा पद (7) है। कुल (42) मिलता है। / The first term is (35) and the second term is (7). The total is (42).
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How many consecutive factors remain in (\frac{(n+3)!}{n!})?
B. (3)
(\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1))। इसलिए तीन क्रमागत गुणक बचते हैं। / (\frac{(n+3)!}{n!}=(n+3)(n+2)(n+1)). Therefore, three consecutive factors remain.
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If \(x=\frac{9!}{7!}\) and \(y=\frac{6!}{4!}\), what is the value of \(\frac{x}{y}\)?
A. \(\frac{12}{5}\)
(x=72) और (y=30), इसलिए \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\)। अनुपात को अंत में सरल करें। / (x=72) and (y=30), so \(\frac{x}{y}=\frac{72}{30}=\frac{12}{5}\). Simplify the ratio at the end.
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What is the simplified form of (\frac{(n+2)!}{n!}+\frac{(n+1)!}{n!})?
A. ((n+1)(n+3))
पहला पद ((n+2)(n+1)) और दूसरा (n+1) है। समान ((n+1)) लेने पर ((n+1)(n+3)) मिलता है। / The first term is ((n+2)(n+1)) and the second is (n+1). Taking common ((n+1)) gives ((n+1)(n+3)).
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What is the value of \(\frac{13!}{11!}-\frac{10!}{8!}\)?
B. (66)
\(\frac{13!}{11!}=13\times12=156\) और \(\frac{10!}{8!}=10\times9=90\), इसलिए अंतर (66) है। पहले दोनों फैक्टोरियल अनुपात अलग-अलग सरल करें। / \(\frac{13!}{11!}=13\times12=156\) and \(\frac{10!}{8!}=10\times9=90\), so the difference is (66). Simplify both factorial ratios separately first.
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If (\frac{(n+2)!}{(n-1)!}=210), what is the value of (n)?
B. (5)
(\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)n)। \(7\times6\times5=210\), इसलिए (n=5) है। / (\frac{(n+2)!}{(n-1)!}=(n+2)(n+1)n). Since \(7\times6\times5=210\), (n=5).
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What is the value of \(\frac{9!-8!}{7!}\)?
B. (64)
अंश (8!(9-1)=8\cdot8!) है। \(\frac{8\cdot8!}{7!}=8\times8=64\) मिलता है। / The numerator is (8!(9-1)=8\cdot8!). Thus \(\frac{8\cdot8!}{7!}=8\times8=64\).
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What is the simplified form of (\frac{(n+4)!}{(n+1)!})?
B. ((n+4)(n+3)(n+2))
((n+4)!=(n+4)(n+3)(n+2)(n+1)!), इसलिए ((n+1)!) कट जाता है। तीन क्रमागत गुणक बचते हैं। / ((n+4)!=(n+4)(n+3)(n+2)(n+1)!), so ((n+1)!) cancels. Three consecutive factors remain.
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