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 431-440 of 609 questions.
If \(\binom{n}{2}=190\), what is the value of (n)?
D. (20)
\(\binom{20}{2}=190\) होता है। इसलिए (n=20) सही है। / \(\binom{20}{2}=190\). Therefore (n=20) is correct.
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If \(\binom{n}{7}=\binom{n}{11}\), what will be the value of (n)?
C. (18)
\(\binom{n}{r}=\binom{n}{s}\) में यहां (r+s=n) होगा। इसलिए (7+11=18) है। / Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (7+11=18).
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If \(\binom{n}{1}+\binom{n}{2}=136\), what is the value of (n)?
B. (16)
यह (\frac{n(n+1)}{2}=136) देता है। (n=16) रखने पर (136) मिलता है। / It gives (\frac{n(n+1)}{2}=136). Putting (n=16) gives (136).
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Using Pascal's identity \(\binom{13}{6}+\binom{13}{7}\) is equal to which expression?
A. \(\binom{14}{7}\)
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{14}{7}\) है। / By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{14}{7}\).
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What is the value of \(\binom{15}{7}-\binom{14}{7}\)?
C. (3003)
\(\binom{15}{7}=6435\) और \(\binom{14}{7}=3432\) हैं। अंतर (3003) है। / \(\binom{15}{7}=6435\) and \(\binom{14}{7}=3432\). The difference is (3003).
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B. (252)
(1) और (2) तय हैं तथा (3), (4) हट गए हैं। बाकी (5) तत्व (10) में से चुने जाएंगे इसलिए \(\binom{10}{5}=252\) है। / The elements (1), (2) are fixed and (3), (4) are excluded. The remaining (5) elements are chosen from (10), so \(\binom{10}{5}=252\).
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A. (672)
(1) तय है इसलिए (5) तत्व (12) में से चुनेंगे। (5), (6) दोनों होने पर \(\binom{10}{3}=120\) घटाने से (792-120=672) मिलता है। / The element (1) is fixed so choose (5) elements from (12). Subtracting \(\binom{10}{3}=120\) cases containing both (5), (6) gives (792-120=672).
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D. (54600)
(6) की (8) गुणज और (40) गैर-गुणज संख्याएं हैं। तरीके \(\binom{8}{4}\binom{40}{2}=54600\) हैं। / There are (8) multiples of (6) and (40) non-multiples. The ways are \(\binom{8}{4}\binom{40}{2}=54600\).
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From numbers (1) to (35), how many ways are there to choose (4) odd and (2) even numbers?
C. (416160)
विषम संख्याएं (18) और सम संख्याएं (17) हैं। तरीके \(\binom{18}{4}\binom{17}{2}=416160\) हैं। / There are (18) odd numbers and (17) even numbers. The ways are \(\binom{18}{4}\binom{17}{2}=416160\).
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A. (6435)
(7) विशेष विद्यार्थियों को हटाने पर (15) विद्यार्थी बचते हैं। इसलिए \(\binom{15}{8}=6435\) तरीके हैं। / After removing (7) special students (15) students remain. Hence there are \(\binom{15}{8}=6435\) ways.
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