(9) शिक्षकों और (11) छात्रों में से (6) लोगों का समूह बनाना है जिसमें कम से कम (3) शिक्षक हों। कितने तरीके हैं?
From (9) teachers and (11) students a group of (6) people is to be formed with at least (3) teachers. How many ways are there?
#combinations
#class11
#hard
A (13860)
B (20790)
C (22260)
D (27132)
Explanation opens after your attempt
Correct Answer
C. (22260)
Step 1
Concept
The number of teachers can be (3), (4), (5), or (6). The sum of all these cases is (22260).
Step 2
Why this answer is correct
The correct answer is C. (22260). The number of teachers can be (3), (4), (5), or (6). The sum of all these cases is (22260).
Step 3
Exam Tip
शिक्षक (3), (4), (5) या (6) हो सकते हैं। इन सभी मामलों का योग (22260) है।
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(18) विद्यार्थियों में से (7) चुनने हैं ताकि (5) विशेष विद्यार्थियों में से कोई भी शामिल न हो। कितने तरीके हैं?
From (18) students (7) are to be selected so that none of (5) special students is included. How many ways are there?
#combinations
#class11
#hard
A (1287)
B (1716)
C (31824)
D (8568)
Explanation opens after your attempt
Step 1
Concept
After removing (5) special students (13) students remain. Hence there are \(\binom{13}{7}=1716\) ways.
Step 2
Why this answer is correct
The correct answer is B. (1716). After removing (5) special students (13) students remain. Hence there are \(\binom{13}{7}=1716\) ways.
Step 3
Exam Tip
(5) विशेष विद्यार्थियों को हटाने पर (13) विद्यार्थी बचते हैं। इसलिए \(\binom{13}{7}=1716\) तरीके हैं।
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(1) से (25) तक की संख्याओं में से (3) विषम और (2) सम संख्या चुनने के कितने तरीके हैं?
From numbers (1) to (25), how many ways are there to choose (3) odd and (2) even numbers?
#combinations
#class11
#hard
A (8580)
B (17160)
C (18876)
D (37752)
Explanation opens after your attempt
Correct Answer
C. (18876)
Step 1
Concept
There are (13) odd numbers and (12) even numbers. The ways are \(\binom{13}{3}\binom{12}{2}=18876\).
Step 2
Why this answer is correct
The correct answer is C. (18876). There are (13) odd numbers and (12) even numbers. The ways are \(\binom{13}{3}\binom{12}{2}=18876\).
Step 3
Exam Tip
विषम संख्याएं (13) और सम संख्याएं (12) हैं। तरीके \(\binom{13}{3}\binom{12}{2}=18876\) हैं।
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(1) से (30) तक की संख्याओं में से (3) ऐसी संख्याएं चुननी हैं जो (3) की गुणज हों और (2) ऐसी संख्याएं जो (3) की गुणज न हों। कितने तरीके हैं?
From numbers (1) to (30), (3) numbers that are multiples of (3) and (2) numbers that are not multiples of (3) are to be selected. How many ways are there?
#combinations
#class11
#hard
A (22800)
B (11400)
C (19000)
D (45600)
Explanation opens after your attempt
Correct Answer
A. (22800)
Step 1
Concept
There are (10) multiples of (3) and (20) non-multiples. The ways are \(\binom{10}{3}\binom{20}{2}=22800\).
Step 2
Why this answer is correct
The correct answer is A. (22800). There are (10) multiples of (3) and (20) non-multiples. The ways are \(\binom{10}{3}\binom{20}{2}=22800\).
Step 3
Exam Tip
(3) की (10) गुणज और (20) गैर-गुणज संख्याएं हैं। तरीके \(\binom{10}{3}\binom{20}{2}=22800\) हैं।
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\(A=\{1,2,3,4,5,6,7,8,9,10,11\}\) के कितने (6)-तत्व उपसमुच्चय (4) और (5) दोनों को साथ शामिल नहीं करते?
How many (6)-element subsets of \(A=\{1,2,3,4,5,6,7,8,9,10,11\}\) do not contain both (4) and (5) together?
#combinations
#class11
#hard
A (280)
B (316)
C (336)
D (462)
Explanation opens after your attempt
Step 1
Concept
Total subsets are \(\binom{11}{6}=462\) and those containing both (4), (5) are \(\binom{9}{4}=126\). Hence (462-126=336).
Step 2
Why this answer is correct
The correct answer is C. (336). Total subsets are \(\binom{11}{6}=462\) and those containing both (4), (5) are \(\binom{9}{4}=126\). Hence (462-126=336).
Step 3
Exam Tip
कुल \(\binom{11}{6}=462\) हैं और (4), (5) दोनों हों तो \(\binom{9}{4}=126\) हैं। इसलिए (462-126=336) है।
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कितने (5)-तत्व उपसमुच्चय \(A=\{1,2,3,4,5,6,7,8,9,10,11,12\}\) से बनाए जा सकते हैं जिनमें (1) और (2) शामिल हों लेकिन (3) शामिल न हो?
How many (5)-element subsets can be formed from \(A=\{1,2,3,4,5,6,7,8,9,10,11,12\}\) that contain (1) and (2) but not (3)?
#combinations
#class11
#hard
A (56)
B (84)
C (120)
D (126)
Explanation opens after your attempt
Step 1
Concept
The elements (1) and (2) are fixed and (3) is excluded. The remaining (3) elements are chosen from (9), so \(\binom{9}{3}=84\).
Step 2
Why this answer is correct
The correct answer is B. (84). The elements (1) and (2) are fixed and (3) is excluded. The remaining (3) elements are chosen from (9), so \(\binom{9}{3}=84\).
Step 3
Exam Tip
(1) और (2) तय हैं और (3) हट गया है। बाकी (3) तत्व (9) में से चुने जाएंगे इसलिए \(\binom{9}{3}=84\) है।
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\(\binom{13}{6}-\binom{12}{6}\) का मान क्या है?
What is the value of \(\binom{13}{6}-\binom{12}{6}\)?
#combinations
#class11
#hard
A (462)
B (715)
C (792)
D (924)
Explanation opens after your attempt
Step 1
Concept
\(\binom{13}{6}=1716\) and \(\binom{12}{6}=924\). The difference is (792).
Step 2
Why this answer is correct
The correct answer is C. (792). \(\binom{13}{6}=1716\) and \(\binom{12}{6}=924\). The difference is (792).
Step 3
Exam Tip
\(\binom{13}{6}=1716\) और \(\binom{12}{6}=924\) हैं। अंतर (792) है।
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\(\binom{11}{4}+\binom{11}{5}\) पास्कल पहचान से किसके बराबर है?
Using Pascal's identity \(\binom{11}{4}+\binom{11}{5}\) is equal to which expression?
#combinations
#class11
#hard
A \(\binom{12}{4}\)
B \(\binom{12}{5}\)
C \(\binom{11}{9}\)
D \(\binom{22}{9}\)
Explanation opens after your attempt
Correct Answer
B. \(\binom{12}{5}\)
Step 1
Concept
By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{12}{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\binom{12}{5}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{12}{5}\).
Step 3
Exam Tip
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{12}{5}\) है।
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यदि \(\binom{n}{1}+\binom{n}{2}=78\) है तो (n) का मान क्या है?
If \(\binom{n}{1}+\binom{n}{2}=78\), what is the value of (n)?
#combinations
#class11
#hard
A (10)
B (11)
C (12)
D (13)
Explanation opens after your attempt
Step 1
Concept
It gives (n+\frac{n(n-1)}{2}=78). Putting (n=12) gives (12+66=78).
Step 2
Why this answer is correct
The correct answer is C. (12). It gives (n+\frac{n(n-1)}{2}=78). Putting (n=12) gives (12+66=78).
Step 3
Exam Tip
यह (n+\frac{n(n-1)}{2}=78) देता है। (n=12) रखने पर (12+66=78) मिलता है।
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यदि \(\binom{n}{5}=\binom{n}{9}\) है तो (n) का मान क्या होगा?
If \(\binom{n}{5}=\binom{n}{9}\), what will be the value of (n)?
#combinations
#class11
#hard
A (12)
B (13)
C (14)
D (15)
Explanation opens after your attempt
Step 1
Concept
Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (5+9=14).
Step 2
Why this answer is correct
The correct answer is C. (14). Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (5+9=14).
Step 3
Exam Tip
\(\binom{n}{r}=\binom{n}{s}\) में यहां (r+s=n) होगा। इसलिए (5+9=14) है।
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यदि \(\binom{n}{2}=120\) है तो (n) का मान क्या है?
If \(\binom{n}{2}=120\), what is the value of (n)?
#combinations
#class11
#hard
A (14)
B (15)
C (16)
D (17)
Explanation opens after your attempt
Step 1
Concept
\(\binom{16}{2}=120\). Therefore (n=16) is correct.
Step 2
Why this answer is correct
The correct answer is C. (16). \(\binom{16}{2}=120\). Therefore (n=16) is correct.
Step 3
Exam Tip
\(\binom{16}{2}=120\) होता है। इसलिए (n=16) सही है।
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(9) अलग-अलग पेन और (8) अलग-अलग पेंसिलों में से (6) वस्तुएं चुननी हैं जिनमें कम से कम (2) पेन और कम से कम (2) पेंसिल हों। कितने तरीके हैं?
From (9) different pens and (8) different pencils (6) objects are to be selected with at least (2) pens and at least (2) pencils. How many ways are there?
#combinations
#class11
#hard
A (9408)
B (10080)
C (10752)
D (12376)
Explanation opens after your attempt
Correct Answer
C. (10752)
Step 1
Concept
The number of pens can be (2), (3), or (4). The total is \(\binom{9}{2}\binom{8}{4}+\binom{9}{3}\binom{8}{3}+\binom{9}{4}\binom{8}{2}=10752\).
Step 2
Why this answer is correct
The correct answer is C. (10752). The number of pens can be (2), (3), or (4). The total is \(\binom{9}{2}\binom{8}{4}+\binom{9}{3}\binom{8}{3}+\binom{9}{4}\binom{8}{2}=10752\).
Step 3
Exam Tip
पेन (2), (3) या (4) हो सकते हैं। कुल \(\binom{9}{2}\binom{8}{4}+\binom{9}{3}\binom{8}{3}+\binom{9}{4}\binom{8}{2}=10752\) है।
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(10) पुरुषों और (8) महिलाओं में से (6) व्यक्तियों की समिति बनानी है जिसमें ठीक (3) पुरुष हों। कितने तरीके हैं?
From (10) men and (8) women a committee of (6) persons is to be formed with exactly (3) men. How many ways are there?
#combinations
#class11
#hard
A (3360)
B (5040)
C (6720)
D (10080)
Explanation opens after your attempt
Step 1
Concept
Exactly (3) men and (3) women are needed. The ways are \(\binom{10}{3}\binom{8}{3}=6720\).
Step 2
Why this answer is correct
The correct answer is C. (6720). Exactly (3) men and (3) women are needed. The ways are \(\binom{10}{3}\binom{8}{3}=6720\).
Step 3
Exam Tip
ठीक (3) पुरुष और (3) महिलाएं चाहिए। तरीके \(\binom{10}{3}\binom{8}{3}=6720\) हैं।
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(15) छात्रों में से (6) छात्रों का चयन करना है। (5) विशेष छात्रों में से कम से कम (2) शामिल हों। कितने तरीके हैं?
From (15) students (6) students are to be selected. At least (2) of (5) special students must be included. How many ways are there?
#combinations
#class11
#hard
A (3535)
B (3745)
C (4265)
D (5005)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{15}{6}=5005\). Removing selections with (0) and (1) special student gives \(5005-\binom{10}{6}-\binom{5}{1}\binom{10}{5}=3535\).
Step 2
Why this answer is correct
The correct answer is A. (3535). Total ways are \(\binom{15}{6}=5005\). Removing selections with (0) and (1) special student gives \(5005-\binom{10}{6}-\binom{5}{1}\binom{10}{5}=3535\).
Step 3
Exam Tip
कुल \(\binom{15}{6}=5005\) हैं। (0) और (1) विशेष वाले चयन हटाने पर \(5005-\binom{10}{6}-\binom{5}{1}\binom{10}{5}=3535\) है।
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(12) विद्यार्थियों में से (6) विद्यार्थियों की टीम बनानी है। (5) विशेष विद्यार्थियों में से ठीक (3) शामिल होने चाहिए। कितने तरीके हैं?
A team of (6) students is to be formed from (12) students. Exactly (3) of (5) special students must be included. How many ways are there?
#combinations
#class11
#hard
A (280)
B (320)
C (350)
D (420)
Explanation opens after your attempt
Step 1
Concept
Choose (3) from (5) special students and (3) from the remaining (7). The ways are \(\binom{5}{3}\binom{7}{3}=350\).
Step 2
Why this answer is correct
The correct answer is C. (350). Choose (3) from (5) special students and (3) from the remaining (7). The ways are \(\binom{5}{3}\binom{7}{3}=350\).
Step 3
Exam Tip
(5) विशेष में से (3) और बाकी (7) में से (3) चुनेंगे। तरीके \(\binom{5}{3}\binom{7}{3}=350\) हैं।
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(11) रंगों में से (5) रंग चुनने हैं लेकिन एक निश्चित रंग अवश्य हो और दूसरा निश्चित रंग न हो। कितने तरीके हैं?
From (11) colors (5) colors are to be selected with one fixed color included and another fixed color excluded. How many ways are there?
#combinations
#class11
#hard
A (70)
B (126)
C (210)
D (252)
Explanation opens after your attempt
Step 1
Concept
One color is fixed and one is removed so choose the remaining (4) colors from (9). The ways are \(\binom{9}{4}=126\).
Step 2
Why this answer is correct
The correct answer is B. (126). One color is fixed and one is removed so choose the remaining (4) colors from (9). The ways are \(\binom{9}{4}=126\).
Step 3
Exam Tip
एक रंग तय है और एक हट गया है इसलिए बाकी (4) रंग (9) में से चुनेंगे। तरीके \(\binom{9}{4}=126\) हैं।
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(13) अलग-अलग उपहारों में से कम से कम (3) और अधिकतम (5) उपहार चुनने के कितने तरीके हैं?
In how many ways can at least (3) and at most (5) gifts be selected from (13) different gifts?
#combinations
#class11
#hard
A (1716)
B (2002)
C (2288)
D (2717)
Explanation opens after your attempt
Step 1
Concept
The selection can be of (3), (4), or (5) gifts. The total is \(\binom{13}{3}+\binom{13}{4}+\binom{13}{5}=2288\).
Step 2
Why this answer is correct
The correct answer is C. (2288). The selection can be of (3), (4), or (5) gifts. The total is \(\binom{13}{3}+\binom{13}{4}+\binom{13}{5}=2288\).
Step 3
Exam Tip
चयन (3), (4) या (5) उपहारों का होगा। कुल \(\binom{13}{3}+\binom{13}{4}+\binom{13}{5}=2288\) है।
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(9) डॉक्टरों और (7) नर्सों में से (6) लोगों की टीम बनानी है जिसमें डॉक्टरों की संख्या नर्सों से अधिक हो। कितने तरीके हैं?
From (9) doctors and (7) nurses a team of (6) people is to be formed with more doctors than nurses. How many ways are there?
#combinations
#class11
#hard
A (2772)
B (3360)
C (3612)
D (4368)
Explanation opens after your attempt
Step 1
Concept
The number of doctors will be (4), (5), or (6). The total is \(\binom{9}{4}\binom{7}{2}+\binom{9}{5}\binom{7}{1}+\binom{9}{6}=3612\).
Step 2
Why this answer is correct
The correct answer is C. (3612). The number of doctors will be (4), (5), or (6). The total is \(\binom{9}{4}\binom{7}{2}+\binom{9}{5}\binom{7}{1}+\binom{9}{6}=3612\).
Step 3
Exam Tip
डॉक्टरों की संख्या (4), (5) या (6) होगी। कुल \(\binom{9}{4}\binom{7}{2}+\binom{9}{5}\binom{7}{1}+\binom{9}{6}=3612\) है।
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(8) अंग्रेजी और (9) हिंदी पुस्तकों में से (5) पुस्तकें चुननी हैं जिनमें दोनों भाषाओं की पुस्तकें हों। कितने तरीके हैं?
From (8) English and (9) Hindi books (5) books are to be selected with books from both languages. How many ways are there?
#combinations
#class11
#hard
A (6006)
B (6088)
C (6132)
D (6188)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{17}{5}=6188\). Removing only English \(\binom{8}{5}=56\) and only Hindi \(\binom{9}{5}=126\) gives (6006).
Step 2
Why this answer is correct
The correct answer is A. (6006). Total ways are \(\binom{17}{5}=6188\). Removing only English \(\binom{8}{5}=56\) and only Hindi \(\binom{9}{5}=126\) gives (6006).
Step 3
Exam Tip
कुल \(\binom{17}{5}=6188\) हैं। केवल अंग्रेजी \(\binom{8}{5}=56\) और केवल हिंदी \(\binom{9}{5}=126\) हटाने पर (6006) मिलते हैं।
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(10) गणित और (7) भौतिकी पुस्तकों में से (6) पुस्तकें चुननी हैं जिनमें कम से कम (3) गणित पुस्तकें हों। कितने तरीके हैं?
From (10) mathematics and (7) physics books (6) books are to be selected with at least (3) mathematics books. How many ways are there?
#combinations
#class11
#hard
A (8820)
B (10080)
C (10584)
D (12376)
Explanation opens after your attempt
Correct Answer
C. (10584)
Step 1
Concept
The cases are (3), (4), (5), and (6) mathematics books. Their correct sum is (10584).
Step 2
Why this answer is correct
The correct answer is C. (10584). The cases are (3), (4), (5), and (6) mathematics books. Their correct sum is (10584).
Step 3
Exam Tip
मामले (3), (4), (5) और (6) गणित पुस्तकों के हैं। इनका सही योग (10584) है।
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(11) व्यक्तियों में से (5) व्यक्तियों की समिति बनानी है और (2) विशेष व्यक्ति साथ में नहीं चुने जा सकते। कितने तरीके होंगे?
A committee of (5) persons is to be formed from (11) persons and (2) special persons cannot be selected together. How many ways are possible?
#combinations
#class11
#hard
A (336)
B (378)
C (420)
D (462)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{11}{5}=462\) and ways with both special persons are \(\binom{9}{3}=84\). Hence (462-84=378).
Step 2
Why this answer is correct
The correct answer is B. (378). Total ways are \(\binom{11}{5}=462\) and ways with both special persons are \(\binom{9}{3}=84\). Hence (462-84=378).
Step 3
Exam Tip
कुल \(\binom{11}{5}=462\) हैं और दोनों विशेष साथ हों तो \(\binom{9}{3}=84\) हैं। इसलिए (462-84=378) तरीके हैं।
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एक लीग में (14) टीमें हैं और हर दो टीमों के बीच (3) मैच खेले जाते हैं। कुल मैच कितने होंगे?
A league has (14) teams and (3) matches are played between every pair of teams. How many matches will be played?
#combinations
#class11
#hard
A (91)
B (182)
C (273)
D (364)
Explanation opens after your attempt
Step 1
Concept
The pairs of teams are \(\binom{14}{2}=91\). With (3) matches for each pair the total is (273).
Step 2
Why this answer is correct
The correct answer is C. (273). The pairs of teams are \(\binom{14}{2}=91\). With (3) matches for each pair the total is (273).
Step 3
Exam Tip
टीमों की जोड़ियां \(\binom{14}{2}=91\) हैं। हर जोड़ी के (3) मैच होने से कुल (273) मैच होंगे।
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(16) बिंदुओं में से (6) बिंदु एक सीध में हैं और अन्य (4) बिंदु भी एक अलग सीध में हैं। कोई अन्य (3) बिंदु एक सीध में नहीं हैं। कितने त्रिभुज बनेंगे?
Among (16) points (6) points are collinear and another (4) points are collinear on a different line. No other (3) points are collinear. How many triangles can be formed?
#combinations
#class11
#hard
A (536)
B (540)
C (556)
D (560)
Explanation opens after your attempt
Step 1
Concept
Total triples are \(\binom{16}{3}=560\). Failed triples are \(\binom{6}{3}+\binom{4}{3}=24\), so (560-24=536).
Step 2
Why this answer is correct
The correct answer is A. (536). Total triples are \(\binom{16}{3}=560\). Failed triples are \(\binom{6}{3}+\binom{4}{3}=24\), so (560-24=536).
Step 3
Exam Tip
कुल \(\binom{16}{3}=560\) त्रिक हैं। असफल त्रिक \(\binom{6}{3}+\binom{4}{3}=24\) हैं इसलिए (560-24=536) है।
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(14) वस्तुओं में से (5) वस्तुएं चुननी हैं और (5) विशेष वस्तुओं में से अधिकतम (1) चुनी जाए। कितने तरीके हैं?
From (14) objects (5) objects are to be selected and at most (1) of (5) special objects is chosen. How many ways are there?
#combinations
#class11
#medium
A (1008)
B (1260)
C (1512)
D (1764)
Explanation opens after your attempt
Step 1
Concept
Either (0) or (1) special object can be chosen. The total is \(\binom{5}{0}\binom{9}{5}+\binom{5}{1}\binom{9}{4}=756\).
Step 2
Why this answer is correct
The correct answer is C. (1512). Either (0) or (1) special object can be chosen. The total is \(\binom{5}{0}\binom{9}{5}+\binom{5}{1}\binom{9}{4}=756\).
Step 3
Exam Tip
विशेष वस्तुओं में से (0) या (1) चुनी जा सकती है। कुल \(\binom{5}{0}\binom{9}{5}+\binom{5}{1}\binom{9}{4}=756\) है।
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(10) खिलाड़ियों में से (4) खिलाड़ियों का चयन करना है। एक कप्तान पहले से तय है और उसे चुना नहीं जाना है। कितने तरीके हैं?
From (10) players (4) players are to be selected. One captain is already fixed and must not be selected. How many ways are there?
#combinations
#class11
#medium
A (84)
B (126)
C (210)
D (504)
Explanation opens after your attempt
Step 1
Concept
After removing the captain (9) players remain. Hence there are \(\binom{9}{4}=126\) ways.
Step 2
Why this answer is correct
The correct answer is B. (126). After removing the captain (9) players remain. Hence there are \(\binom{9}{4}=126\) ways.
Step 3
Exam Tip
कप्तान को हटाने पर (9) खिलाड़ी बचते हैं। इसलिए \(\binom{9}{4}=126\) तरीके हैं।
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(9) अलग-अलग खिलौनों में से विषम संख्या में खिलौने चुनने के कितने तरीके हैं?
In how many ways can an odd number of toys be selected from (9) different toys?
#combinations
#class11
#medium
A (128)
B (255)
C (256)
D (512)
Explanation opens after your attempt
Step 1
Concept
The number of odd selections is \(2^{9-1}=256\). Even and odd selections are equal.
Step 2
Why this answer is correct
The correct answer is C. (256). The number of odd selections is \(2^{9-1}=256\). Even and odd selections are equal.
Step 3
Exam Tip
विषम चयन की संख्या \(2^{9-1}=256\) होती है। सम और विषम चयन बराबर होते हैं।
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(10) अलग-अलग सिक्कों में से सम संख्या में सिक्के चुनने के कितने तरीके हैं?
In how many ways can an even number of coins be selected from (10) different coins?
#combinations
#class11
#medium
A (256)
B (512)
C (1023)
D (1024)
Explanation opens after your attempt
Step 1
Concept
An even selection means (0), (2), (4), (6), (8), or (10) coins. The total number is \(2^{10-1}=512\).
Step 2
Why this answer is correct
The correct answer is B. (512). An even selection means (0), (2), (4), (6), (8), or (10) coins. The total number is \(2^{10-1}=512\).
Step 3
Exam Tip
सम संख्या का चयन (0), (2), (4), (6), (8), (10) सिक्कों का होगा। कुल संख्या \(2^{10-1}=512\) है।
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(12) अलग-अलग कार्डों में से (5) कार्ड चुनने हैं जिनमें दो निश्चित कार्डों में से ठीक एक कार्ड हो। कितने तरीके हैं?
From (12) distinct cards (5) cards are to be selected containing exactly one of two fixed cards. How many ways are there?
#combinations
#class11
#medium
A (336)
B (420)
C (504)
D (672)
Explanation opens after your attempt
Step 1
Concept
Choose (1) of the two fixed cards and (4) cards from the remaining (10). The ways are \(\binom{2}{1}\binom{10}{4}=420\).
Step 2
Why this answer is correct
The correct answer is B. (420). Choose (1) of the two fixed cards and (4) cards from the remaining (10). The ways are \(\binom{2}{1}\binom{10}{4}=420\).
Step 3
Exam Tip
दो निश्चित कार्डों में से (1) चुनें और बाकी (4) कार्ड (10) में से चुनें। तरीके \(\binom{2}{1}\binom{10}{4}=420\) हैं।
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(7) वरिष्ठ और (9) कनिष्ठ कर्मचारियों में से (5) लोगों की समिति बनानी है जिसमें वरिष्ठों की संख्या कनिष्ठों से कम हो। कितने तरीके हैं?
From (7) senior and (9) junior employees a committee of (5) is to be formed with fewer seniors than juniors. How many ways are there?
#combinations
#class11
#medium
A (1911)
B (2016)
C (2111)
D (4368)
Explanation opens after your attempt
Step 1
Concept
The number of seniors can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{9}{5}+\binom{7}{1}\binom{9}{4}+\binom{7}{2}\binom{9}{3}=2772\).
Step 2
Why this answer is correct
The correct answer is B. (2016). The number of seniors can be (0), (1), or (2). The total is \(\binom{7}{0}\binom{9}{5}+\binom{7}{1}\binom{9}{4}+\binom{7}{2}\binom{9}{3}=2772\).
Step 3
Exam Tip
वरिष्ठों की संख्या (0), (1) या (2) हो सकती है। कुल \(\binom{7}{0}\binom{9}{5}+\binom{7}{1}\binom{9}{4}+\binom{7}{2}\binom{9}{3}=2772\) है।
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(10) प्रश्नों में से (6) प्रश्न चुनने हैं और अंतिम (4) प्रश्नों में से कम से कम (2) प्रश्न चुनने हैं। कितने चयन होंगे?
From (10) questions (6) are to be selected and at least (2) of the last (4) questions must be selected. How many selections are there?
#combinations
#class11
#medium
A (140)
B (170)
C (185)
D (210)
Explanation opens after your attempt
Step 1
Concept
The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{6}{4}+\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{6}{2}=185\).
Step 2
Why this answer is correct
The correct answer is C. (185). The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{6}{4}+\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{6}{2}=185\).
Step 3
Exam Tip
मामले अंतिम (4) में से (2), (3) या (4) चुनने के हैं। कुल \(\binom{4}{2}\binom{6}{4}+\binom{4}{3}\binom{6}{3}+\binom{4}{4}\binom{6}{2}=185\) है।
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