(5) अलग-अलग कार्डों में से किसी भी संख्या में कार्ड चुनने के कितने तरीके हैं?
In how many ways can any number of cards be selected from (5) different cards?
#combinations
#class11
#easy
A (5)
B (25)
C (31)
D (32)
Explanation opens after your attempt
Step 1
Concept
Each card may be selected or not selected. Hence the total ways are \(2^5=32\).
Step 2
Why this answer is correct
The correct answer is D. (32). Each card may be selected or not selected. Hence the total ways are \(2^5=32\).
Step 3
Exam Tip
हर कार्ड चुना या न चुना जा सकता है। इसलिए कुल तरीके \(2^5=32\) हैं।
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(6) अलग-अलग सिक्कों में से कम से कम (1) सिक्का चुनने के कितने तरीके हैं?
In how many ways can at least (1) coin be selected from (6) different coins?
#combinations
#class11
#easy
A (31)
B (32)
C (63)
D (64)
Explanation opens after your attempt
Step 1
Concept
The ways for at least (1) selection are \(2^6-1=63\). Do not forget to subtract the empty selection.
Step 2
Why this answer is correct
The correct answer is C. (63). The ways for at least (1) selection are \(2^6-1=63\). Do not forget to subtract the empty selection.
Step 3
Exam Tip
कम से कम (1) चयन के तरीके \(2^6-1=63\) हैं। खाली चयन को घटाना न भूलें।
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पास्कल पहचान के अनुसार \(\binom{5}{2}+\binom{5}{3}\) किसके बराबर है?
By Pascal's identity, \(\binom{5}{2}+\binom{5}{3}\) is equal to which expression?
#combinations
#class11
#easy
A \(\binom{6}{3}\)
B \(\binom{6}{2}\)
C \(\binom{5}{5}\)
D \(\binom{10}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\binom{6}{3}\)
Step 1
Concept
Pascal's identity is \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{6}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{6}{3}\). Pascal's identity is \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{6}{3}\).
Step 3
Exam Tip
पास्कल पहचान \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) है। इसलिए उत्तर \(\binom{6}{3}\) है।
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\(\binom{6}{2}+\binom{6}{4}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{6}{2}+\binom{6}{4}\).
#combinations
#class11
#easy
A (15)
B (24)
C (30)
D (45)
Explanation opens after your attempt
Step 1
Concept
\(\binom{6}{2}=15\) and \(\binom{6}{4}=15\). The total is (30).
Step 2
Why this answer is correct
The correct answer is C. (30). \(\binom{6}{2}=15\) and \(\binom{6}{4}=15\). The total is (30).
Step 3
Exam Tip
\(\binom{6}{2}=15\) और \(\binom{6}{4}=15\) हैं। कुल (30) मिलेगा।
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यदि \(\binom{n}{1}+\binom{n}{0}=16\) है तो (n) का मान क्या है?
If \(\binom{n}{1}+\binom{n}{0}=16\), what is the value of (n)?
#combinations
#class11
#easy
A (14)
B (15)
C (16)
D (17)
Explanation opens after your attempt
Step 1
Concept
It gives (n+1=16). Therefore (n=15).
Step 2
Why this answer is correct
The correct answer is B. (15). It gives (n+1=16). Therefore (n=15).
Step 3
Exam Tip
यह (n+1=16) देता है। इसलिए (n=15) होगा।
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यदि \(\binom{n}{2}=21\) है तो (n) का मान क्या है?
If \(\binom{n}{2}=21\), what is the value of (n)?
#combinations
#class11
#easy
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
\(\binom{7}{2}=21\). Hence (n=7) is correct.
Step 2
Why this answer is correct
The correct answer is B. (7). \(\binom{7}{2}=21\). Hence (n=7) is correct.
Step 3
Exam Tip
\(\binom{7}{2}=21\) होता है। इसलिए (n=7) सही है।
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\(\binom{7}{2}+\binom{7}{5}\) का मान क्या है?
What is the value of \(\binom{7}{2}+\binom{7}{5}\)?
#combinations
#class11
#easy
A (21)
B (35)
C (42)
D (70)
Explanation opens after your attempt
Step 1
Concept
\(\binom{7}{2}=21\) and \(\binom{7}{5}=21\). Therefore the total is (42).
Step 2
Why this answer is correct
The correct answer is C. (42). \(\binom{7}{2}=21\) and \(\binom{7}{5}=21\). Therefore the total is (42).
Step 3
Exam Tip
\(\binom{7}{2}=21\) और \(\binom{7}{5}=21\) हैं। इसलिए कुल (42) है।
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\(\binom{9}{5}\) किसके बराबर है?
\(\binom{9}{5}\) is equal to which expression?
#combinations
#class11
#easy
A \(\binom{9}{4}\)
B \(\binom{9}{3}\)
C \(\binom{5}{9}\)
D \(\binom{4}{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\binom{9}{4}\)
Step 1
Concept
Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (9-5=4).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{9}{4}\). Because \(\binom{n}{r}=\binom{n}{n-r}\). Here (9-5=4).
Step 3
Exam Tip
क्योंकि \(\binom{n}{r}=\binom{n}{n-r}\) होता है। यहां (9-5=4) है।
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\(\binom{12}{4}\) का मान क्या है?
What is the value of \(\binom{12}{4}\)?
#combinations
#class11
#easy
A (48)
B (220)
C (495)
D (11880)
Explanation opens after your attempt
Step 1
Concept
\(\binom{12}{4}=495\). Dividing by (4!) is necessary in calculation.
Step 2
Why this answer is correct
The correct answer is C. (495). \(\binom{12}{4}=495\). Dividing by (4!) is necessary in calculation.
Step 3
Exam Tip
\(\binom{12}{4}=495\) होता है। गणना में (4!) से भाग देना जरूरी है।
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\(\binom{13}{2}\) का मान ज्ञात कीजिए।
Find the value of \(\binom{13}{2}\).
#combinations
#class11
#easy
A (26)
B (39)
C (78)
D (156)
Explanation opens after your attempt
Step 1
Concept
\(\binom{13}{2}=\frac{13\cdot12}{2}=78\). For choosing two objects, take half of the product.
Step 2
Why this answer is correct
The correct answer is C. (78). \(\binom{13}{2}=\frac{13\cdot12}{2}=78\). For choosing two objects, take half of the product.
Step 3
Exam Tip
\(\binom{13}{2}=\frac{13\cdot12}{2}=78\) है। दो वस्तुओं के चयन में आधा गुणनफल लें।
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\(\binom{10}{10}\) का मान क्या है?
What is the value of \(\binom{10}{10}\)?
#combinations
#class11
#easy
A (0)
B (1)
C (10)
D (100)
Explanation opens after your attempt
Step 1
Concept
There is exactly one way to choose all objects. Thus \(\binom{10}{10}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is exactly one way to choose all objects. Thus \(\binom{10}{10}=1\).
Step 3
Exam Tip
सभी वस्तुएं चुनने का एक ही तरीका होता है। अतः \(\binom{10}{10}=1\) है।
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\(\binom{15}{0}\) का मान क्या होगा?
What will be the value of \(\binom{15}{0}\)?
#combinations
#class11
#easy
A (0)
B (1)
C (15)
D (30)
Explanation opens after your attempt
Step 1
Concept
There is one way to choose zero objects. Hence \(\binom{15}{0}=1\).
Step 2
Why this answer is correct
The correct answer is B. (1). There is one way to choose zero objects. Hence \(\binom{15}{0}=1\).
Step 3
Exam Tip
शून्य वस्तु चुनने का एक तरीका होता है। इसलिए \(\binom{15}{0}=1\) है।
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\(\binom{14}{1}\) का मान क्या है?
What is the value of \(\binom{14}{1}\)?
#combinations
#class11
#easy
A (1)
B (13)
C (14)
D (28)
Explanation opens after your attempt
Step 1
Concept
\(\binom{n}{1}=n\). Therefore \(\binom{14}{1}=14\).
Step 2
Why this answer is correct
The correct answer is C. (14). \(\binom{n}{1}=n\). Therefore \(\binom{14}{1}=14\).
Step 3
Exam Tip
\(\binom{n}{1}=n\) होता है। इसलिए \(\binom{14}{1}=14\) है।
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(8) विद्यार्थियों में से (4) चुनने हैं और दो विशेष विद्यार्थी साथ-साथ नहीं चुने जाने चाहिए। कितने तरीके हैं?
From (8) students, (4) are to be selected and two special students should not be selected together. How many ways are there?
#combinations
#class11
#easy
A (55)
B (60)
C (65)
D (70)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{8}{4}=70\), and ways with both special students are \(\binom{6}{2}=15\). Thus (70-15=55) ways remain.
Step 2
Why this answer is correct
The correct answer is A. (55). Total ways are \(\binom{8}{4}=70\), and ways with both special students are \(\binom{6}{2}=15\). Thus (70-15=55) ways remain.
Step 3
Exam Tip
कुल \(\binom{8}{4}=70\) और दोनों विशेष साथ हों तो \(\binom{6}{2}=15\) तरीके हैं। इसलिए (70-15=55) तरीके बचेंगे।
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(7) विद्यार्थियों में से (3) विद्यार्थियों को चुनना है और दो विशेष विद्यार्थी दोनों साथ में चुने जाने चाहिए। कितने तरीके हैं?
From (7) students, (3) students are to be selected and two special students must be selected together. How many ways are there?
#combinations
#class11
#easy
A (3)
B (5)
C (10)
D (21)
Explanation opens after your attempt
Step 1
Concept
The two special students are already selected. The third student is chosen from the remaining (5) in \(\binom{5}{1}=5\) ways.
Step 2
Why this answer is correct
The correct answer is B. (5). The two special students are already selected. The third student is chosen from the remaining (5) in \(\binom{5}{1}=5\) ways.
Step 3
Exam Tip
दो विशेष विद्यार्थी पहले से चुने गए हैं। तीसरा विद्यार्थी शेष (5) में से \(\binom{5}{1}=5\) तरीकों से चुनेगा।
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(9) सदस्यों में से (3) सदस्यों की समिति बनानी है जिसमें एक निश्चित सदस्य शामिल न हो। कितने तरीके हैं?
A committee of (3) members is to be formed from (9) members excluding one fixed member. How many ways are there?
#combinations
#class11
#easy
A (56)
B (84)
C (168)
D (24)
Explanation opens after your attempt
Step 1
Concept
After excluding one member, (8) members remain. Therefore there are \(\binom{8}{3}=56\) ways.
Step 2
Why this answer is correct
The correct answer is A. (56). After excluding one member, (8) members remain. Therefore there are \(\binom{8}{3}=56\) ways.
Step 3
Exam Tip
एक सदस्य हटाने पर (8) सदस्य बचते हैं। इसलिए \(\binom{8}{3}=56\) तरीके हैं।
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(10) सदस्यों में से (4) सदस्यों की समिति बनानी है जिसमें एक निश्चित सदस्य अवश्य हो। कितने तरीके हैं?
A committee of (4) members is to be formed from (10) members with one fixed member included. How many ways are there?
#combinations
#class11
#easy
A (84)
B (126)
C (210)
D (5040)
Explanation opens after your attempt
Step 1
Concept
The fixed member is already selected, so choose the remaining (3) from (9). The ways are \(\binom{9}{3}=84\).
Step 2
Why this answer is correct
The correct answer is A. (84). The fixed member is already selected, so choose the remaining (3) from (9). The ways are \(\binom{9}{3}=84\).
Step 3
Exam Tip
निश्चित सदस्य पहले से चुना है इसलिए बाकी (3) सदस्य (9) में से चुनेंगे। तरीके \(\binom{9}{3}=84\) हैं।
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(6) सदस्यों में से अध्यक्ष या सचिव बनाए बिना (3) सदस्यों की समिति कितने तरीकों से बनेगी?
In how many ways can a committee of (3) members be formed from (6) members without assigning president or secretary?
#combinations
#class11
#easy
A (18)
B (20)
C (120)
D (216)
Explanation opens after your attempt
Step 1
Concept
No posts are assigned, so it is a combination. \(\binom{6}{3}=20\).
Step 2
Why this answer is correct
The correct answer is B. (20). No posts are assigned, so it is a combination. \(\binom{6}{3}=20\).
Step 3
Exam Tip
पद नहीं दिए गए हैं इसलिए यह संयोजन है। \(\binom{6}{3}=20\) होगा।
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(16) टीमों के टूर्नामेंट में प्रत्येक टीम हर दूसरी टीम से एक बार खेले तो कुल मैच कितने होंगे?
In a tournament of (16) teams, if each team plays every other team once, how many matches will be played?
#combinations
#class11
#easy
A (120)
B (160)
C (240)
D (256)
Explanation opens after your attempt
Step 1
Concept
Each match is a pair of (2) teams. Hence there will be \(\binom{16}{2}=120\) matches.
Step 2
Why this answer is correct
The correct answer is A. (120). Each match is a pair of (2) teams. Hence there will be \(\binom{16}{2}=120\) matches.
Step 3
Exam Tip
प्रत्येक मैच (2) टीमों की जोड़ी है। इसलिए \(\binom{16}{2}=120\) मैच होंगे।
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(14) व्यक्तियों में से (2) व्यक्तियों के हाथ मिलाने की कुल संख्या क्या होगी?
What is the total number of handshakes among (14) persons?
#combinations
#class11
#easy
A (28)
B (91)
C (182)
D (196)
Explanation opens after your attempt
Step 1
Concept
Each handshake is a pair of (2) persons. The total is \(\binom{14}{2}=91\).
Step 2
Why this answer is correct
The correct answer is B. (91). Each handshake is a pair of (2) persons. The total is \(\binom{14}{2}=91\).
Step 3
Exam Tip
प्रत्येक हाथ मिलाना (2) व्यक्तियों की जोड़ी है। कुल \(\binom{14}{2}=91\) होंगे।
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(9) बिंदुओं में से कोई (3) एक सीध में नहीं हैं। कितने वृत्त बन सकते हैं?
There are (9) points, no (3) of which are collinear. How many circles can be formed?
#combinations
#class11
#easy
A (27)
B (72)
C (84)
D (126)
Explanation opens after your attempt
Step 1
Concept
A circle is determined by (3) points. Hence \(\binom{9}{3}=84\) is correct.
Step 2
Why this answer is correct
The correct answer is C. (84). A circle is determined by (3) points. Hence \(\binom{9}{3}=84\) is correct.
Step 3
Exam Tip
एक वृत्त (3) बिंदुओं से निर्धारित होता है। इसलिए \(\binom{9}{3}=84\) सही है।
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(8) बिंदुओं में से कोई (3) एक सीध में नहीं हैं। इनसे कितने त्रिभुज बन सकते हैं?
There are (8) points, no (3) of which are collinear. How many triangles can be formed?
#combinations
#class11
#easy
A (24)
B (56)
C (112)
D (336)
Explanation opens after your attempt
Step 1
Concept
A triangle is formed by choosing (3) points. Hence \(\binom{8}{3}=56\).
Step 2
Why this answer is correct
The correct answer is B. (56). A triangle is formed by choosing (3) points. Hence \(\binom{8}{3}=56\).
Step 3
Exam Tip
त्रिभुज के लिए (3) बिंदु चुने जाते हैं। इसलिए \(\binom{8}{3}=56\) है।
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(13) बिंदुओं में से (2) बिंदु चुनकर कितने रेखाखंड बनाए जा सकते हैं?
How many line segments can be formed by choosing (2) points from (13) points?
#combinations
#class11
#easy
A (26)
B (39)
C (78)
D (156)
Explanation opens after your attempt
Step 1
Concept
A line segment needs (2) points. Therefore \(\binom{13}{2}=78\).
Step 2
Why this answer is correct
The correct answer is C. (78). A line segment needs (2) points. Therefore \(\binom{13}{2}=78\).
Step 3
Exam Tip
एक रेखाखंड के लिए (2) बिंदु चाहिए। अतः \(\binom{13}{2}=78\) होगा।
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(4) गणित और (5) भौतिकी पुस्तकों में से (1) गणित और (2) भौतिकी पुस्तकें कितने तरीकों से चुनी जा सकती हैं?
From (4) mathematics and (5) physics books, in how many ways can (1) mathematics book and (2) physics books be chosen?
#combinations
#class11
#easy
A (20)
B (30)
C (40)
D (60)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). The multiplication rule applies to selections from different subjects.
Step 2
Why this answer is correct
The correct answer is C. (40). The ways are \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\). The multiplication rule applies to selections from different subjects.
Step 3
Exam Tip
तरीके \(\binom{4}{1}\binom{5}{2}=4\cdot10=40\) हैं। अलग विषयों के चयन में गुणा नियम लगता है।
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(7) पुरुषों और (3) महिलाओं में से (1) पुरुष और (2) महिलाएं कितने तरीकों से चुनी जा सकती हैं?
From (7) men and (3) women, in how many ways can (1) man and (2) women be selected?
#combinations
#class11
#easy
A (10)
B (21)
C (42)
D (63)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\). Write each condition as a separate combination.
Step 2
Why this answer is correct
The correct answer is B. (21). The ways are \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\). Write each condition as a separate combination.
Step 3
Exam Tip
तरीके \(\binom{7}{1}\binom{3}{2}=7\cdot3=21\) हैं। प्रत्येक शर्त को अलग संयोजन से लिखें।
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(6) लाल और (5) नीली गेंदों में से (2) लाल और (1) नीली गेंद कितने तरीकों से चुनी जा सकती है?
From (6) red and (5) blue balls, in how many ways can (2) red and (1) blue ball be selected?
#combinations
#class11
#easy
A (30)
B (45)
C (60)
D (75)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\). Select separately when colors are different.
Step 2
Why this answer is correct
The correct answer is D. (75). The ways are \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\). Select separately when colors are different.
Step 3
Exam Tip
तरीके \(\binom{6}{2}\binom{5}{1}=15\cdot5=75\) हैं। रंग अलग हों तो चयन अलग-अलग करें।
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(5) लड़कों और (4) लड़कियों में से (2) लड़के और (2) लड़कियां कितने तरीकों से चुनी जा सकती हैं?
From (5) boys and (4) girls, in how many ways can (2) boys and (2) girls be selected?
#combinations
#class11
#easy
A (20)
B (40)
C (60)
D (120)
Explanation opens after your attempt
Step 1
Concept
The ways are \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\). Multiply selections from different groups.
Step 2
Why this answer is correct
The correct answer is C. (60). The ways are \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\). Multiply selections from different groups.
Step 3
Exam Tip
तरीके \(\binom{5}{2}\binom{4}{2}=10\cdot6=60\) हैं। अलग वर्गों से चयन में गुणा करते हैं।
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(10) प्रश्नों में से (7) प्रश्न चुनने के कितने तरीके हैं?
How many ways are there to select (7) questions from (10) questions?
#combinations
#class11
#easy
A (120)
B (210)
C (5040)
D (70)
Explanation opens after your attempt
Step 1
Concept
\(\binom{10}{7}=\binom{10}{3}=120\). Complementary selection makes calculation faster.
Step 2
Why this answer is correct
The correct answer is A. (120). \(\binom{10}{7}=\binom{10}{3}=120\). Complementary selection makes calculation faster.
Step 3
Exam Tip
\(\binom{10}{7}=\binom{10}{3}=120\) है। पूरक चयन से गणना जल्दी होती है।
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(12) पेन में से (3) पेन चुनने के कितने तरीके हैं?
How many ways are there to choose (3) pens from (12) pens?
#combinations
#class11
#easy
A (36)
B (132)
C (220)
D (440)
Explanation opens after your attempt
Step 1
Concept
The number of ways is \(\binom{12}{3}=220\). The order of pens is not considered in selection.
Step 2
Why this answer is correct
The correct answer is C. (220). The number of ways is \(\binom{12}{3}=220\). The order of pens is not considered in selection.
Step 3
Exam Tip
तीन पेन चुनने के तरीके \(\binom{12}{3}=220\) हैं। चयन में पेन का क्रम नहीं देखा जाता।
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(11) खिलाड़ियों में से (4) खिलाड़ियों का अभ्यास समूह कितने तरीकों से चुना जा सकता है?
In how many ways can a practice group of (4) players be selected from (11) players?
#combinations
#class11
#easy
A (44)
B (165)
C (330)
D (7920)
Explanation opens after your attempt
Step 1
Concept
This is selection, so the value is \(\binom{11}{4}=330\). The correct option is (330).
Step 2
Why this answer is correct
The correct answer is B. (165). This is selection, so the value is \(\binom{11}{4}=330\). The correct option is (330).
Step 3
Exam Tip
यह चयन है इसलिए \(\binom{11}{4}=330\) नहीं बल्कि सही मान \(\binom{11}{4}=330\) है। विकल्पों में सही उत्तर (330) है।
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