class11 se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
1558 questions tagged with class11.
In how many ways can any number of cards be selected from (5) different cards?
D. (32)
हर कार्ड चुना या न चुना जा सकता है। इसलिए कुल तरीके \(2^5=32\) हैं। / Each card may be selected or not selected. Hence the total ways are \(2^5=32\).
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In how many ways can at least (1) coin be selected from (6) different coins?
C. (63)
कम से कम (1) चयन के तरीके \(2^6-1=63\) हैं। खाली चयन को घटाना न भूलें। / The ways for at least (1) selection are \(2^6-1=63\). Do not forget to subtract the empty selection.
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By Pascal's identity, \(\binom{5}{2}+\binom{5}{3}\) is equal to which expression?
A. \(\binom{6}{3}\)
पास्कल पहचान \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) है। इसलिए उत्तर \(\binom{6}{3}\) है। / Pascal's identity is \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{6}{3}\).
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Find the value of \(\binom{6}{2}+\binom{6}{4}\).
C. (30)
\(\binom{6}{2}=15\) और \(\binom{6}{4}=15\) हैं। कुल (30) मिलेगा। / \(\binom{6}{2}=15\) and \(\binom{6}{4}=15\). The total is (30).
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If \(\binom{n}{1}+\binom{n}{0}=16\), what is the value of (n)?
B. (15)
यह (n+1=16) देता है। इसलिए (n=15) होगा। / It gives (n+1=16). Therefore (n=15).
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If \(\binom{n}{2}=21\), what is the value of (n)?
B. (7)
\(\binom{7}{2}=21\) होता है। इसलिए (n=7) सही है। / \(\binom{7}{2}=21\). Hence (n=7) is correct.
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What is the value of \(\binom{7}{2}+\binom{7}{5}\)?
C. (42)
\(\binom{7}{2}=21\) और \(\binom{7}{5}=21\) हैं। इसलिए कुल (42) है। / \(\binom{7}{2}=21\) and \(\binom{7}{5}=21\). Therefore the total is (42).
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\(\binom{9}{5}\) is equal to which expression?
A. \(\binom{9}{4}\)
क्योंकि \(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (9-5=4) है। / Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (9-5=4).
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What is the value of \(\binom{12}{4}\)?
C. (495)
\(\binom{12}{4}=495\) होता है। गणना में (4!) से भाग देना जरूरी है। / \(\binom{12}{4}=495\). Dividing by (4!) is necessary in calculation.
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Find the value of \(\binom{13}{2}\).
C. (78)
\(\binom{13}{2}=\frac{13\cdot12}{2}=78\) है। दो वस्तुओं के चयन में आधा गुणनफल लें। / \(\binom{13}{2}=\frac{13\cdot12}{2}=78\). For choosing two objects, take half of the product.
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What is the value of \(\binom{10}{10}\)?
B. (1)
सभी वस्तुएं चुनने का एक ही तरीका होता है। अतः \(\binom{10}{10}=1\) है। / There is exactly one way to choose all objects. Thus \(\binom{10}{10}=1\).
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What will be the value of \(\binom{15}{0}\)?
B. (1)
शून्य वस्तु चुनने का एक तरीका होता है। इसलिए \(\binom{15}{0}=1\) है। / There is one way to choose zero objects. Hence \(\binom{15}{0}=1\).
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What is the value of \(\binom{14}{1}\)?
C. (14)
\(\binom{n}{1}=n\) होता है। इसलिए \(\binom{14}{1}=14\) है। / \(\binom{n}{1}=n\). Therefore \(\binom{14}{1}=14\).
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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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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. (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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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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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. (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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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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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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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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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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C. (40)
तरीके \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\) हैं। अलग विषयों के चयन में गुणा नियम लगता है। / The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). The multiplication rule applies to selections from different subjects.
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From (7) men and (3) women, in how many ways can (1) man and (2) women be selected?
B. (21)
तरीके \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\) हैं। प्रत्येक शर्त को अलग संयोजन से लिखें। / The ways are \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\). Write each condition as a separate combination.
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From (6) red and (5) blue balls, in how many ways can (2) red and (1) blue ball be selected?
D. (75)
तरीके \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\) हैं। रंग अलग हों तो चयन अलग-अलग करें। / The ways are \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\). Select separately when colors are different.
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From (5) boys and (4) girls, in how many ways can (2) boys and (2) girls be selected?
C. (60)
तरीके \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\) हैं। अलग वर्गों से चयन में गुणा करते हैं। / The ways are \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\). Multiply selections from different groups.
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How many ways are there to select (7) questions from (10) questions?
A. (120)
\(\binom{10}{7}=\binom{10}{3}=120\) है। पूरक चयन से गणना जल्दी होती है। / \(\binom{10}{7}=\binom{10}{3}=120\). Complementary selection makes calculation faster.
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How many ways are there to choose (3) pens from (12) pens?
C. (220)
तीन पेन चुनने के तरीके \(\binom{12}{3}=220\) हैं। चयन में पेन का क्रम नहीं देखा जाता। / The number of ways is \(\binom{12}{3}=220\). The order of pens is not considered in selection.
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In how many ways can a practice group of (4) players be selected from (11) players?
B. (165)
यह चयन है इसलिए \(\binom{11}{4}=330\) नहीं बल्कि सही मान \(\binom{11}{4}=330\) है। विकल्पों में सही उत्तर (330) है। / This is selection, so the value is \(\binom{11}{4}=330\). The correct option is (330).
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