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 381-390 of 609 questions.
If \(\binom{n}{2}=153\), what is the value of (n)?
C. (18)
\(\binom{18}{2}=153\) होता है। इसलिए (n=18) सही है। / \(\binom{18}{2}=153\). Therefore (n=18) is correct.
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If \(\binom{n}{6}=\binom{n}{10}\), what will be the value of (n)?
C. (16)
\(\binom{n}{r}=\binom{n}{s}\) में यहां (r+s=n) होगा। इसलिए (6+10=16) है। / Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (6+10=16).
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If \(\binom{n}{1}+\binom{n}{2}=105\), what is the value of (n)?
C. (14)
यह (n+\frac{n(n-1)}{2}=105) देता है। (n=14) रखने पर (14+91=105) मिलता है। / It gives (n+\frac{n(n-1)}{2}=105). Putting (n=14) gives (14+91=105).
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Using Pascal's identity \(\binom{12}{5}+\binom{12}{6}\) is equal to which expression?
A. \(\binom{13}{6}\)
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{13}{6}\) है। / By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{13}{6}\).
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What is the value of \(\binom{14}{7}-\binom{13}{7}\)?
B. (1716)
\(\binom{14}{7}=3432\) और \(\binom{13}{7}=1716\) हैं। अंतर (1716) है। / \(\binom{14}{7}=3432\) and \(\binom{13}{7}=1716\). The difference is (1716).
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B. (126)
(1) और (2) तय हैं तथा (3), (4) हट गए हैं। बाकी (4) तत्व (9) में से चुने जाएंगे इसलिए \(\binom{9}{4}=126\) है। / The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (4) elements are chosen from (9), so \(\binom{9}{4}=126\).
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B. (378)
(1) तय है इसलिए (5) तत्व (11) में से चुनेंगे। (5), (6) दोनों होने पर \(\binom{9}{3}=84\) घटाने से (462-84=378) मिलता है। / The element (1) is fixed so choose (5) elements from (11). Subtracting \(\binom{9}{3}=84\) cases containing both (5), (6) gives (462-84=378).
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B. (29484)
(4) की (9) गुणज और (27) गैर-गुणज संख्याएं हैं। तरीके \(\binom{9}{3}\binom{27}{2}=29484\) हैं। / There are (9) multiples of (4) and (27) non-multiples. The ways are \(\binom{9}{3}\binom{27}{2}=29484\).
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From numbers (1) to (30), how many ways are there to choose (3) odd and (3) even numbers?
C. (207025)
विषम संख्याएं (15) और सम संख्याएं (15) हैं। तरीके \(\binom{15}{3}\binom{15}{3}=207025\) हैं। / There are (15) odd numbers and (15) even numbers. The ways are \(\binom{15}{3}\binom{15}{3}=207025\).
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B. (3432)
(6) विशेष विद्यार्थियों को हटाने पर (14) विद्यार्थी बचते हैं। इसलिए \(\binom{14}{7}=3432\) तरीके हैं। / After removing (6) special students (14) students remain. Hence there are \(\binom{14}{7}=3432\) ways.
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