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 141-150 of 609 questions.
What is the value of \(\binom{12}{2}+\binom{12}{10}\)?
C. (132)
\(\binom{12}{2}=66\) और \(\binom{12}{10}=66\) हैं। कुल (132) होगा। / \(\binom{12}{2}=66\) and \(\binom{12}{10}=66\). The total is (132).
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Using Pascal's identity, \(\binom{6}{2}+\binom{6}{3}\) is equal to which expression?
C. \(\binom{7}{4}\)
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{7}{3}\) नहीं, सही \(\binom{7}{3}\) है। / By Pascal's identity, \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the correct expression is \(\binom{7}{3}\).
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If \(\binom{n}{2}=45\), what is the value of (n)?
C. (10)
\(\binom{10}{2}=45\) होता है। इसलिए (n=10) सही है। / \(\binom{10}{2}=45\). Therefore (n=10) is correct.
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If \(\binom{n}{1}+\binom{n}{0}=21\), what will be the value of (n)?
B. (20)
यह (n+1=21) देता है। इसलिए (n=20) होगा। / It gives (n+1=21). Therefore (n=20).
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If \(\binom{n}{0}+\binom{n}{n}=2\), what type of identity does it show?
A. सीमा पहचानBoundary identity
क्योंकि \(\binom{n}{0}=1\) और \(\binom{n}{n}=1\) होते हैं। यह संयोजन की सीमा पहचान है। / Because \(\binom{n}{0}=1\) and \(\binom{n}{n}=1\). This is a boundary identity of combinations.
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B. (10)
विशेष मिठाई पहले से चुनी है। बाकी (2) मिठाइयां (5) में से \(\binom{5}{2}=10\) तरीकों से चुनेंगे। / The special sweet is already selected. Choose the remaining (2) sweets from (5) in \(\binom{5}{2}=10\) ways.
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A. (35)
टूटे खिलौने को हटाने पर (7) खिलौने बचते हैं। इसलिए \(\binom{7}{3}=35\) तरीके हैं। / After removing the broken toy, (7) toys remain. Hence there are \(\binom{7}{3}=35\) ways.
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C. (60)
ठीक (2) हरी और (1) नीली चाहिए। तरीके \(\binom{6}{2}\binom{4}{1}=15\cdot4=60\) हैं। / Exactly (2) green and (1) blue are needed. The ways are \(\binom{6}{2}\binom{4}{1}=15\cdot4=60\).
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D. (60)
तरीके \(\binom{3}{1}\binom{4}{1}\binom{5}{1}=60\) हैं। तीन अलग समूहों के चयन को गुणा करें। / The ways are \(\binom{3}{1}\binom{4}{1}\binom{5}{1}=60\). Multiply selections from the three different groups.
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A. (45)
पहले (2) प्रश्न तय हैं इसलिए बाकी (2) प्रश्न (10) में से चुनेंगे। तरीके \(\binom{10}{2}=45\) हैं। / The first (2) questions are fixed, so choose the remaining (2) from (10). The ways are \(\binom{10}{2}=45\).
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