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 411-420 of 591 questions.
What is the value of \( \frac{14!}{12!\cdot2!}+\frac{13!}{11!\cdot2!} \)?
B. (169)
दोनों पद क्रमशः (91) और (78) हैं, इसलिए योग (169) है। जोड़ने से पहले हर पद अलग निकालें। / The two terms are (91) and (78), so the sum is (169). Find each term separately before adding.
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What is the value of \( \frac{18!}{16!\cdot2!}-\frac{17!}{15!\cdot2!} \)?
C. (17)
पहला पद (153) और दूसरा (136) है, इसलिए अंतर (17) है। निकट फैक्टोरियल पदों का अंतर सावधानी से घटाएं। / The first term is (153) and the second is (136), so the difference is (17). Subtract nearby factorial terms carefully.
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What is the value of \( \frac{10!}{6!\cdot4!}+\frac{10!}{7!\cdot3!} \)?
D. (330)
दोनों पद (210) और (120) हैं, इसलिए योग (330) है। संयोजन जैसे रूपों में छोटे हर को सही रखें। / The two terms are (210) and (120), so the sum is (330). In combination-like forms, keep the small denominator correct.
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If ( \frac{(2n+1)!}{(2n-1)!}=306 ), what is the value of (n)?
B. (8)
यह ((2n+1)(2n)=306) है और \(17\cdot16=272\) नहीं; सही समीकरण के लिए कोई पूर्णांक (n) नहीं मिलेगा। / It is ((2n+1)(2n)=306), and no integer (n) satisfies it. Check whether the matched consecutive factors fit the form.
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If ( \frac{(2n+1)!}{(2n-1)!}=272 ), what is the value of (n)?
C. (8)
यह ((2n+1)(2n)=272) है और \(17\cdot16=272\), इसलिए (2n=16) और (n=8)। पहले लगातार गुणकों को पहचानें। / It is ((2n+1)(2n)=272), and \(17\cdot16=272\), so (2n=16) and (n=8). Identify the consecutive factors first.
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What is ( \frac{(3n+1)!}{(3n-1)!} ) equal to?
A. ( (3n+1)(3n) )
((3n+1)!=(3n+1)(3n)(3n-1)!), इसलिए दो गुणक बचते हैं। सूचक वाले पद को एक इकाई की तरह संभालें। / ((3n+1)!=(3n+1)(3n)(3n-1)!), so two factors remain. Handle the indexed term as one unit.
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If ( \frac{(3n+1)!}{(3n-1)!}=156 ), what is the value of (n)?
B. (4)
यह ((3n+1)(3n)=156) है और \(13\cdot12=156\), इसलिए (3n=12) और (n=4)। रूप पहचानकर मान रखें। / It is ((3n+1)(3n)=156), and \(13\cdot12=156\), so (3n=12) and (n=4). Identify the form and substitute.
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What is the simplified form of ( \frac{(n!)2}{(n-2)!(n+2)!} )?
B. ( \frac{n(n-1)}{(n+1)(n+2)} )
( \frac{n!}{(n-2)!}=n(n-1) ) और ( \frac{n!}{(n+2)!}=\frac{1}{(n+2)(n+1)} ), इसलिए उत्तर मिलता है। अनुपात को दो भागों में तोड़ें। / ( \frac{n!}{(n-2)!}=n(n-1) ) and ( \frac{n!}{(n+2)!}=\frac{1}{(n+2)(n+1)} ), so the answer follows. Break the ratio into two parts.
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If ( \frac{(n+3)!}{n!}-\frac{(n+2)!}{(n-1)!}=270 ), what is the value of (n)?
B. (8)
सरल रूप (3(n+2)(n+1)) है, इसलिए ((n+2)(n+1)=90)। \(10\cdot9=90\), अतः (n=8)। / The simplified form is (3(n+2)(n+1)), so ((n+2)(n+1)=90). Since \(10\cdot9=90\), (n=8).
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How many zeros will be at the end of (20!)?
B. (4)
अंतिम शून्यों की संख्या \( \left\lfloor\frac{20}{5}\right\rfloor=4 \) है। शून्य गिनने में (5) के गुणकों को गिनें। / The number of ending zeros is \( \left\lfloor\frac{20}{5}\right\rfloor=4 \). To count zeros, count multiples of (5).
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