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 231-240 of 609 questions.
If \(\binom{n}{1}+\binom{n}{2}=45\), what is the value of (n)?
B. (9)
यह (n+\frac{n(n-1)}{2}=45) देता है। (n=9) रखने पर (9+36=45) मिलता है। / It gives (n+\frac{n(n-1)}{2}=45). Putting (n=9) gives (9+36=45).
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Using Pascal's identity \(\binom{9}{4}+\binom{9}{5}\) is equal to which expression?
A. \(\binom{10}{5}\)
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{10}{5}\) है। / By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{10}{5}\).
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What is the value of \(\binom{11}{5}-\binom{10}{5}\)?
B. (210)
\(\binom{11}{5}=462\) और \(\binom{10}{5}=252\) हैं। अंतर (210) है। / \(\binom{11}{5}=462\) and \(\binom{10}{5}=252\). The difference is (210).
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How many (4)-element subsets can be formed from \(A=\{1,2,3,4,5,6,7,8,9\}\) that contain (1)?
B. (56)
(1) पहले से चुना है इसलिए बाकी (3) तत्व (8) में से चुने जाएंगे। तरीकों की संख्या \(\binom{8}{3}=56\) है। / The element (1) is already chosen so the remaining (3) elements are chosen from (8). The number of ways is \(\binom{8}{3}=56\).
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A. (91)
कुल \(\binom{9}{5}=126\) हैं और (2), (3) दोनों हों तो \(\binom{7}{3}=35\) हैं। इसलिए (126-35=91) है। / Total subsets are \(\binom{9}{5}=126\) and those containing both (2), (3) are \(\binom{7}{3}=35\). Hence (126-35=91).
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From numbers (1) to (18), how many ways are there to choose (4) even numbers?
B. (126)
(1) से (18) तक (9) सम संख्याएं हैं। इसलिए \(\binom{9}{4}=126\) तरीके हैं। / There are (9) even numbers from (1) to (18). Hence there are \(\binom{9}{4}=126\) ways.
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From numbers (1) to (20), how many ways are there to choose (2) odd and (2) even numbers?
A. (2025)
विषम संख्याएं (10) और सम संख्याएं (10) हैं। तरीके \(\binom{10}{2}\binom{10}{2}=2025\) हैं। / There are (10) odd numbers and (10) even numbers. The ways are \(\binom{10}{2}\binom{10}{2}=2025\).
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B. (462)
(4) विशेष विद्यार्थियों को हटाने पर (11) विद्यार्थी बचते हैं। इसलिए \(\binom{11}{6}=462\) तरीके हैं। / After removing (4) special students (11) students remain. Hence there are \(\binom{11}{6}=462\) ways.
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B. (3213)
कुल \(\binom{16}{5}=4368\) हैं। (0) शिक्षक और (1) शिक्षक के मामले हटाने पर \(4368-\binom{9}{5}-\binom{7}{1}\binom{9}{4}=3213\) है। / Total ways are \(\binom{16}{5}=4368\). Removing the cases of (0) teacher and (1) teacher gives \(4368-\binom{9}{5}-\binom{7}{1}\binom{9}{4}=3213\).
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B. (1260)
कुल \(\binom{13}{5}=1287\) हैं। सभी सफेद \(\binom{6}{5}=6\) और सभी काली \(\binom{7}{5}=21\) हटाने पर (1260) मिलते हैं। / Total ways are \(\binom{13}{5}=1287\). Removing all-white \(\binom{6}{5}=6\) and all-black \(\binom{7}{5}=21\) gives (1260).
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