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 61-70 of 600 questions.
How many line segments can be formed by choosing (2) points from (13) points?
C. (78)
एक रेखाखंड के लिए (2) बिंदु चाहिए। अतः \(\binom{13}{2}=78\) होगा। / A line segment needs (2) points. Therefore \(\binom{13}{2}=78\).
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There are (8) points, no (3) of which are collinear. How many triangles can be formed?
B. (56)
त्रिभुज के लिए (3) बिंदु चुने जाते हैं। इसलिए \(\binom{8}{3}=56\) है। / A triangle is formed by choosing (3) points. Hence \(\binom{8}{3}=56\).
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There are (9) points, no (3) of which are collinear. How many circles can be formed?
C. (84)
एक वृत्त (3) बिंदुओं से निर्धारित होता है। इसलिए \(\binom{9}{3}=84\) सही है। / A circle is determined by (3) points. Hence \(\binom{9}{3}=84\) is correct.
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What is the total number of handshakes among (14) persons?
B. (91)
प्रत्येक हाथ मिलाना (2) व्यक्तियों की जोड़ी है। कुल \(\binom{14}{2}=91\) होंगे। / Each handshake is a pair of (2) persons. The total is \(\binom{14}{2}=91\).
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A. (120)
प्रत्येक मैच (2) टीमों की जोड़ी है। इसलिए \(\binom{16}{2}=120\) मैच होंगे। / Each match is a pair of (2) teams. Hence there will be \(\binom{16}{2}=120\) matches.
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B. (20)
पद नहीं दिए गए हैं इसलिए यह संयोजन है। \(\binom{6}{3}=20\) होगा। / No posts are assigned, so it is a combination. \(\binom{6}{3}=20\).
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A. (84)
निश्चित सदस्य पहले से चुना है इसलिए बाकी (3) सदस्य (9) में से चुनेंगे। तरीके \(\binom{9}{3}=84\) हैं। / The fixed member is already selected, so choose the remaining (3) from (9). The ways are \(\binom{9}{3}=84\).
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A. (56)
एक सदस्य हटाने पर (8) सदस्य बचते हैं। इसलिए \(\binom{8}{3}=56\) तरीके हैं। / After excluding one member, (8) members remain. Therefore there are \(\binom{8}{3}=56\) ways.
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B. (5)
दो विशेष विद्यार्थी पहले से चुने गए हैं। तीसरा विद्यार्थी शेष (5) में से \(\binom{5}{1}=5\) तरीकों से चुनेगा। / The two special students are already selected. The third student is chosen from the remaining (5) in \(\binom{5}{1}=5\) ways.
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A. (55)
कुल \(\binom{8}{4}=70\) और दोनों विशेष साथ हों तो \(\binom{6}{2}=15\) तरीके हैं। इसलिए (70-15=55) तरीके बचेंगे। / Total ways are \(\binom{8}{4}=70\), and ways with both special students are \(\binom{6}{2}=15\). Thus (70-15=55) ways remain.
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