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 81-90 of 609 questions.
Find the value of \(\binom{13}{2}\).
C. (78)
\(\binom{13}{2}=\frac{13\cdot12}{2}=78\) है। दो वस्तुओं के चयन में आधा गुणनफल लें। / \(\binom{13}{2}=\frac{13\cdot12}{2}=78\). For choosing two objects, take half of the product.
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What is the value of \(\binom{12}{4}\)?
C. (495)
\(\binom{12}{4}=495\) होता है। गणना में (4!) से भाग देना जरूरी है। / \(\binom{12}{4}=495\). Dividing by (4!) is necessary in calculation.
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\(\binom{9}{5}\) is equal to which expression?
A. \(\binom{9}{4}\)
क्योंकि \(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (9-5=4) है। / Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (9-5=4).
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What is the value of \(\binom{7}{2}+\binom{7}{5}\)?
C. (42)
\(\binom{7}{2}=21\) और \(\binom{7}{5}=21\) हैं। इसलिए कुल (42) है। / \(\binom{7}{2}=21\) and \(\binom{7}{5}=21\). Therefore the total is (42).
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If \(\binom{n}{2}=21\), what is the value of (n)?
B. (7)
\(\binom{7}{2}=21\) होता है। इसलिए (n=7) सही है। / \(\binom{7}{2}=21\). Hence (n=7) is correct.
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If \(\binom{n}{1}+\binom{n}{0}=16\), what is the value of (n)?
B. (15)
यह (n+1=16) देता है। इसलिए (n=15) होगा। / It gives (n+1=16). Therefore (n=15).
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Find the value of \(\binom{6}{2}+\binom{6}{4}\).
C. (30)
\(\binom{6}{2}=15\) और \(\binom{6}{4}=15\) हैं। कुल (30) मिलेगा। / \(\binom{6}{2}=15\) and \(\binom{6}{4}=15\). The total is (30).
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By Pascal's identity, \(\binom{5}{2}+\binom{5}{3}\) is equal to which expression?
A. \(\binom{6}{3}\)
पास्कल पहचान \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) है। इसलिए उत्तर \(\binom{6}{3}\) है। / Pascal's identity is \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{6}{3}\).
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In how many ways can at least (1) coin be selected from (6) different coins?
C. (63)
कम से कम (1) चयन के तरीके \(2^6-1=63\) हैं। खाली चयन को घटाना न भूलें। / The ways for at least (1) selection are \(2^6-1=63\). Do not forget to subtract the empty selection.
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In how many ways can any number of cards be selected from (5) different cards?
D. (32)
हर कार्ड चुना या न चुना जा सकता है। इसलिए कुल तरीके \(2^5=32\) हैं। / Each card may be selected or not selected. Hence the total ways are \(2^5=32\).
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