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 511-520 of 600 questions.
A. (155)
कुल \(^{11}C_{3}=165\) से कोलिनियर \(^{5}C_{3}=10\) घटाएं। उत्तर (155) है। / Subtract collinear choices \(^{5}C_{3}=10\) from total \(^{11}C_{3}=165\). The answer is (155).
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How many diagonals are there in a convex polygon with (10) sides?
B. (35)
दो शीर्ष चुनने के कुल तरीके \(^{10}C_{2}\) हैं। भुजाएं (10) घटाने पर विकर्ण (35) मिलते हैं। / Total ways to choose two vertices are \(^{10}C_{2}\). Subtract (10) sides to get (35) diagonals.
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How many quadrilaterals can be formed from a convex polygon with (12) sides?
C. (495)
किसी भी (4) शीर्षों से एक चतुर्भुज बनता है। इसलिए संख्या \(^{12}C_{4}=495\) है। / Any (4) vertices form one quadrilateral. Thus the number is \(^{12}C_{4}=495\).
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A. (1589)
कुल \(^{13}C_{6}\) से (0) और (1) गेंदबाज वाले मामले घटाएं। (1716-7-120=1589)। / Subtract cases with (0) and (1) bowler from \(^{13}C_{6}\). (1716-7-120=1589).
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C. (310)
कुल \(^{11}C_{4}=330\) से केवल भारतीय \(^{6}C_{4}=15\) और केवल विदेशी \(^{5}C_{4}=5\) हटाएं। उत्तर (310)। / Subtract all-Indian \(^{6}C_{4}=15\) and all-foreign \(^{5}C_{4}=5\) from total \(^{11}C_{4}=330\). Answer (310).
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A. (875)
पहले (5) में से (3,4,5) चुनकर बाकी (10) से पूर्ति करें। \(^{5}C_{3}{}^{10}C_{7}+^{5}C_{4}{}^{10}C_{6}+^{5}C_{5}{}^{10}C_{5}=875\)। / Choose (3,4,5) from the first (5) and complete from the remaining (10). \(^{5}C_{3}{}^{10}C_{7}+^{5}C_{4}{}^{10}C_{6}+^{5}C_{5}{}^{10}C_{5}=875\).
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D. (945)
मामले (3,5), (4,4), और (5,3) हैं। योग \(^{6}C_{3}{}^{7}C_{5}+^{6}C_{4}{}^{7}C_{4}+^{6}C_{5}{}^{7}C_{3}=945\)। / Cases are (3,5), (4,4), and (5,3). Sum \(^{6}C_{3}{}^{7}C_{5}+^{6}C_{4}{}^{7}C_{4}+^{6}C_{5}{}^{7}C_{3}=945\).
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B. (21)
विशेष अभ्यर्थी को निश्चित मानें और बाकी (2) को (7) में से चुनें। संख्या \(^{7}C_{2}=21\)। / Fix the particular candidate and choose remaining (2) from (7). The number is \(^{7}C_{2}=21\).
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A. (105)
कुल \(^{9}C_{4}=126\) हैं। दोनों विरोधी साथ होने पर \(^{7}C_{2}=21\) हटाएं, उत्तर (105)। / Total selections are \(^{9}C_{4}=126\). Remove \(^{7}C_{2}=21\) selections containing both rivals, answer (105).
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B. (60)
ठीक (2) भौतिकी के लिए \(^{4}C_{2}\) और (3) गणित के लिए \(^{5}C_{3}\)। गुणन से (60) मिलता है। / For exactly (2) physics books use \(^{4}C_{2}\), and for (3) mathematics books use \(^{5}C_{3}\). Product is (60).
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