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 331-340 of 609 questions.
If \(\binom{n}{2}=120\), what is the value of (n)?
C. (16)
\(\binom{16}{2}=120\) होता है। इसलिए (n=16) सही है। / \(\binom{16}{2}=120\). Therefore (n=16) is correct.
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If \(\binom{n}{5}=\binom{n}{9}\), what will be the value of (n)?
C. (14)
\(\binom{n}{r}=\binom{n}{s}\) में यहां (r+s=n) होगा। इसलिए (5+9=14) है। / Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (5+9=14).
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If \(\binom{n}{1}+\binom{n}{2}=78\), what is the value of (n)?
C. (12)
यह (n+\frac{n(n-1)}{2}=78) देता है। (n=12) रखने पर (12+66=78) मिलता है। / It gives (n+\frac{n(n-1)}{2}=78). Putting (n=12) gives (12+66=78).
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Using Pascal's identity \(\binom{11}{4}+\binom{11}{5}\) is equal to which expression?
B. \(\binom{12}{5}\)
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{12}{5}\) है। / By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{12}{5}\).
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What is the value of \(\binom{13}{6}-\binom{12}{6}\)?
C. (792)
\(\binom{13}{6}=1716\) और \(\binom{12}{6}=924\) हैं। अंतर (792) है। / \(\binom{13}{6}=1716\) and \(\binom{12}{6}=924\). The difference is (792).
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B. (84)
(1) और (2) तय हैं और (3) हट गया है। बाकी (3) तत्व (9) में से चुने जाएंगे इसलिए \(\binom{9}{3}=84\) है। / The elements (1) and (2) are fixed and (3) is excluded. The remaining (3) elements are chosen from (9), so \(\binom{9}{3}=84\).
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C. (336)
कुल \(\binom{11}{6}=462\) हैं और (4), (5) दोनों हों तो \(\binom{9}{4}=126\) हैं। इसलिए (462-126=336) है। / Total subsets are \(\binom{11}{6}=462\) and those containing both (4), (5) are \(\binom{9}{4}=126\). Hence (462-126=336).
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A. (22800)
(3) की (10) गुणज और (20) गैर-गुणज संख्याएं हैं। तरीके \(\binom{10}{3}\binom{20}{2}=22800\) हैं। / There are (10) multiples of (3) and (20) non-multiples. The ways are \(\binom{10}{3}\binom{20}{2}=22800\).
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From numbers (1) to (25), how many ways are there to choose (3) odd and (2) even numbers?
C. (18876)
विषम संख्याएं (13) और सम संख्याएं (12) हैं। तरीके \(\binom{13}{3}\binom{12}{2}=18876\) हैं। / There are (13) odd numbers and (12) even numbers. The ways are \(\binom{13}{3}\binom{12}{2}=18876\).
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B. (1716)
(5) विशेष विद्यार्थियों को हटाने पर (13) विद्यार्थी बचते हैं। इसलिए \(\binom{13}{7}=1716\) तरीके हैं। / After removing (5) special students (13) students remain. Hence there are \(\binom{13}{7}=1716\) ways.
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