Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 161-170 of 590 questions.
What is the value of \(\frac{7!-6!}{6!}\)?
B. (6)
\(7!=7\times6!\), इसलिए (\frac{7!-6!}{6!}=\frac{6!(7-1)}{6!}=6)। समान फैक्टोरियल बाहर लें। / Since \(7!=7\times6!\), (\frac{7!-6!}{6!}=\frac{6!(7-1)}{6!}=6). Take the common factorial out.
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What is the simplified form of (\frac{(n+3)!}{n!})?
A. ((n+3)(n+2)(n+1))
((n+3)!=(n+3)(n+2)(n+1)n!), इसलिए (n!) कट जाता है। क्रम से तीन गुणक बचते हैं। / ((n+3)!=(n+3)(n+2)(n+1)n!), so (n!) cancels. Three consecutive factors remain.
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If (\frac{(n+2)!}{n!}=30), what is the value of (n)?
B. (4)
(\frac{(n+2)!}{n!}=(n+2)(n+1))। \(6\times5=30\), इसलिए (n=4)। / (\frac{(n+2)!}{n!}=(n+2)(n+1)). Since \(6\times5=30\), (n=4).
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What is the value of \(\frac{9!}{6!,3!}-\frac{7!}{5!,2!}\)?
A. (63)
\(\frac{9!}{6!,3!}=84\) और \(\frac{7!}{5!,2!}=21\), इसलिए अंतर (63) है। दोनों संयोजन-जैसे पद अलग निकालें। / \(\frac{9!}{6!,3!}=84\) and \(\frac{7!}{5!,2!}=21\), so the difference is (63). Evaluate both combination-like terms separately.
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What is the value of \(\frac{5!+3!}{4!}\)?
C. \(\frac{21}{4}\)
अंश (120+6=126) है और (4!=24)। इसलिए मान \(\frac{126}{24}=\frac{21}{4}\) है। / The numerator is (120+6=126) and (4!=24). Hence the value is \(\frac{126}{24}=\frac{21}{4}\).
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What is the value of \(\frac{8!}{4!}-\frac{7!}{4!}\)?
B. (1470)
(\frac{8!-7!}{4!}=\frac{7!(8-1)}{4!}=7\times7\times6\times5=1470)। समान हर होने पर अंश मिलाकर देखें। / (\frac{8!-7!}{4!}=\frac{7!(8-1)}{4!}=7\times7\times6\times5=1470). With a common denominator, combine the numerator.
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What is the value of \(\frac{11!}{9!,2!}\)?
B. (55)
\(\frac{11!}{9!,2!}=\frac{11\times10}{2}=55\)। बड़े फैक्टोरियल को (9!) तक लिखें। / \(\frac{11!}{9!,2!}=\frac{11\times10}{2}=55\). Write the larger factorial up to (9!).
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Which expression is equal to \(\frac{12!}{9!}\)?
B. \(12\times11\times10\)
\(\frac{12!}{9!}=12\times11\times10\)। हर में (9!) होने से (9!) तक का भाग कट जाता है। / \(\frac{12!}{9!}=12\times11\times10\). Since the denominator is (9!), the part up to (9!) cancels.
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What is the simplified form of (\frac{(n+1)!-n!}{n!})?
B. (n)
((n+1)!=(n+1)n!), इसलिए अंश (n!{(n+1)-1}=nn!) है। भाग देने पर (n) मिलता है। / Since ((n+1)!=(n+1)n!), the numerator is (n!{(n+1)-1}=nn!). Dividing gives (n).
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If (\frac{n!}{(n-1)!}+2=9), what is the value of (n)?
C. (7)
(\frac{n!}{(n-1)!}=n), इसलिए (n+2=9)। अतः (n=7) है। / (\frac{n!}{(n-1)!}=n), so (n+2=9). Therefore, (n=7).
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