यदि \(\binom{n}{3}=\binom{n}{5}\) है तो (n) का मान क्या होगा?
If \(\binom{n}{3}=\binom{n}{5}\), what will be the value of (n)?
#combinations
#class11
#medium
A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
In \(\binom{n}{r}=\binom{n}{s}\) usually (r+s=n). Hence (3+5=8).
Step 2
Why this answer is correct
The correct answer is C. (8). In \(\binom{n}{r}=\binom{n}{s}\) usually (r+s=n). Hence (3+5=8).
Step 3
Exam Tip
\(\binom{n}{r}=\binom{n}{s}\) में सामान्यतः (r+s=n) होता है। इसलिए (3+5=8) है।
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\(\binom{n}{2}=66\) हो तो (n) का मान क्या है?
If \(\binom{n}{2}=66\), what is the value of (n)?
#combinations
#class11
#medium
A (10)
B (11)
C (12)
D (13)
Explanation opens after your attempt
Step 1
Concept
\(\binom{12}{2}=66\). Therefore (n=12) is correct.
Step 2
Why this answer is correct
The correct answer is C. (12). \(\binom{12}{2}=66\). Therefore (n=12) is correct.
Step 3
Exam Tip
\(\binom{12}{2}=66\) होता है। इसलिए (n=12) सही है।
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(6) अलग-अलग पेन और (5) अलग-अलग पेंसिलों में से (4) वस्तुएं चुननी हैं जिनमें कम से कम (1) पेन और (1) पेंसिल हो। कितने तरीके हैं?
From (6) different pens and (5) different pencils (4) objects are to be selected with at least (1) pen and (1) pencil. How many ways are there?
#combinations
#class11
#medium
A (285)
B (300)
C (305)
D (330)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{11}{4}=330\). Removing only pens \(\binom{6}{4}=15\) and only pencils \(\binom{5}{4}=5\) gives (310) ways.
Step 2
Why this answer is correct
The correct answer is C. (305). Total ways are \(\binom{11}{4}=330\). Removing only pens \(\binom{6}{4}=15\) and only pencils \(\binom{5}{4}=5\) gives (310) ways.
Step 3
Exam Tip
कुल \(\binom{11}{4}=330\) हैं। केवल पेन \(\binom{6}{4}=15\) और केवल पेंसिल \(\binom{5}{4}=5\) हटाने पर (310) तरीके मिलते हैं।
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(7) पुरुषों और (6) महिलाओं में से (5) व्यक्तियों की समिति बनानी है जिसमें ठीक (3) पुरुष हों। कितने तरीके हैं?
From (7) men and (6) women a committee of (5) persons is to be formed with exactly (3) men. How many ways are there?
#combinations
#class11
#medium
A (420)
B (525)
C (630)
D (735)
Explanation opens after your attempt
Step 1
Concept
Exactly (3) men and (2) women are needed. The ways are \(\binom{7}{3}\binom{6}{2}=525\).
Step 2
Why this answer is correct
The correct answer is B. (525). Exactly (3) men and (2) women are needed. The ways are \(\binom{7}{3}\binom{6}{2}=525\).
Step 3
Exam Tip
ठीक (3) पुरुष और (2) महिलाएं चाहिए। तरीके \(\binom{7}{3}\binom{6}{2}=525\) हैं।
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(12) छात्रों में से (5) छात्रों का चयन करना है। (4) विशेष छात्रों में से कम से कम (1) शामिल हो। कितने तरीके हैं?
From (12) students (5) students are to be selected. At least (1) of (4) special students must be included. How many ways are there?
#combinations
#class11
#medium
A (736)
B (760)
C (770)
D (792)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{12}{5}=792\) and ways with no special student are \(\binom{8}{5}=56\). Hence (792-56=736).
Step 2
Why this answer is correct
The correct answer is A. (736). Total ways are \(\binom{12}{5}=792\) and ways with no special student are \(\binom{8}{5}=56\). Hence (792-56=736).
Step 3
Exam Tip
कुल \(\binom{12}{5}=792\) हैं और कोई विशेष न हो तो \(\binom{8}{5}=56\) हैं। इसलिए (792-56=736) है।
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(9) विद्यार्थियों में से (4) विद्यार्थियों की टीम बनानी है। (3) विशेष विद्यार्थियों में से ठीक (2) शामिल होने चाहिए। कितने तरीके हैं?
A team of (4) students is to be formed from (9) students. Exactly (2) of (3) special students must be included. How many ways are there?
#combinations
#class11
#medium
A (36)
B (45)
C (54)
D (63)
Explanation opens after your attempt
Step 1
Concept
Choose (2) from (3) special students and (2) from the remaining (6). The ways are \(\binom{3}{2}\binom{6}{2}=45\).
Step 2
Why this answer is correct
The correct answer is B. (45). Choose (2) from (3) special students and (2) from the remaining (6). The ways are \(\binom{3}{2}\binom{6}{2}=45\).
Step 3
Exam Tip
(3) विशेष में से (2) और बाकी (6) में से (2) चुनेंगे। तरीके \(\binom{3}{2}\binom{6}{2}=45\) हैं।
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(8) रंगों में से (3) रंग चुनने हैं लेकिन एक निश्चित रंग अवश्य हो और दूसरा निश्चित रंग न हो। कितने तरीके हैं?
From (8) colors (3) colors are to be selected with one fixed color included and another fixed color excluded. How many ways are there?
#combinations
#class11
#medium
A (15)
B (20)
C (21)
D (35)
Explanation opens after your attempt
Step 1
Concept
One color is fixed and one is removed so choose the remaining (2) colors from (6). The ways are \(\binom{6}{2}=15\).
Step 2
Why this answer is correct
The correct answer is A. (15). One color is fixed and one is removed so choose the remaining (2) colors from (6). The ways are \(\binom{6}{2}=15\).
Step 3
Exam Tip
एक रंग तय है और एक हट गया है इसलिए बाकी (2) रंग (6) में से चुनेंगे। तरीके \(\binom{6}{2}=15\) हैं।
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(10) अलग-अलग उपहारों में से कम से कम (2) और अधिकतम (4) उपहार चुनने के कितने तरीके हैं?
In how many ways can at least (2) and at most (4) gifts be selected from (10) different gifts?
#combinations
#class11
#medium
A (330)
B (375)
C (385)
D (420)
Explanation opens after your attempt
Step 1
Concept
The selection can be of (2), (3), or (4) gifts. The total is \(\binom{10}{2}+\binom{10}{3}+\binom{10}{4}=375\).
Step 2
Why this answer is correct
The correct answer is C. (385). The selection can be of (2), (3), or (4) gifts. The total is \(\binom{10}{2}+\binom{10}{3}+\binom{10}{4}=375\).
Step 3
Exam Tip
चयन (2), (3) या (4) उपहारों का होगा। कुल \(\binom{10}{2}+\binom{10}{3}+\binom{10}{4}=375\) है।
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(6) डॉक्टरों और (4) नर्सों में से (4) लोगों की टीम बनानी है जिसमें डॉक्टरों की संख्या नर्सों से अधिक हो। कितने तरीके हैं?
From (6) doctors and (4) nurses a team of (4) people is to be formed with more doctors than nurses. How many ways are there?
#combinations
#class11
#medium
A (95)
B (105)
C (115)
D (125)
Explanation opens after your attempt
Step 1
Concept
The cases are (3) doctors (1) nurse and (4) doctors. The total is \(\binom{6}{3}\binom{4}{1}+\binom{6}{4}=95\).
Step 2
Why this answer is correct
The correct answer is B. (105). The cases are (3) doctors (1) nurse and (4) doctors. The total is \(\binom{6}{3}\binom{4}{1}+\binom{6}{4}=95\).
Step 3
Exam Tip
मामले (3) डॉक्टर (1) नर्स और (4) डॉक्टर हैं। कुल \(\binom{6}{3}\binom{4}{1}+\binom{6}{4}=95\) है।
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(5) अंग्रेजी और (7) हिंदी पुस्तकों में से (3) पुस्तकें चुननी हैं जिनमें दोनों विषयों की पुस्तकें हों। कितने तरीके हैं?
From (5) English and (7) Hindi books (3) books are to be selected with books from both subjects. How many ways are there?
#combinations
#class11
#medium
A (175)
B (185)
C (190)
D (210)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{12}{3}=220\). Removing only English \(\binom{5}{3}=10\) and only Hindi \(\binom{7}{3}=35\) gives (175) ways.
Step 2
Why this answer is correct
The correct answer is B. (185). Total ways are \(\binom{12}{3}=220\). Removing only English \(\binom{5}{3}=10\) and only Hindi \(\binom{7}{3}=35\) gives (175) ways.
Step 3
Exam Tip
कुल \(\binom{12}{3}=220\) हैं। केवल अंग्रेजी \(\binom{5}{3}=10\) और केवल हिंदी \(\binom{7}{3}=35\) हटाने पर (175) तरीके मिलते हैं।
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(7) गणित और (6) भौतिकी पुस्तकों में से (4) पुस्तकें चुननी हैं जिनमें कम से कम (2) गणित पुस्तकें हों। कितने तरीके हैं?
From (7) mathematics and (6) physics books (4) books are to be selected with at least (2) mathematics books. How many ways are there?
#combinations
#class11
#medium
A (455)
B (490)
C (560)
D (595)
Explanation opens after your attempt
Step 1
Concept
The cases are (2), (3), and (4) mathematics books. The total is \(\binom{7}{2}\binom{6}{2}+\binom{7}{3}\binom{6}{1}+\binom{7}{4}=560\).
Step 2
Why this answer is correct
The correct answer is D. (595). The cases are (2), (3), and (4) mathematics books. The total is \(\binom{7}{2}\binom{6}{2}+\binom{7}{3}\binom{6}{1}+\binom{7}{4}=560\).
Step 3
Exam Tip
मामले (2), (3) और (4) गणित पुस्तकों के हैं। कुल \(\binom{7}{2}\binom{6}{2}+\binom{7}{3}\binom{6}{1}+\binom{7}{4}=560\) है।
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(8) व्यक्तियों में से (3) व्यक्तियों की समिति बनानी है लेकिन (2) विशेष व्यक्ति साथ में नहीं चुने जा सकते। कितने तरीके होंगे?
A committee of (3) persons is to be formed from (8) persons but (2) special persons cannot be selected together. How many ways are possible?
#combinations
#class11
#medium
A (36)
B (42)
C (48)
D (56)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{8}{3}=56\) and if both special persons are together the third person is chosen in (6) ways. Hence the valid count is (50).
Step 2
Why this answer is correct
The correct answer is C. (48). Total ways are \(\binom{8}{3}=56\) and if both special persons are together the third person is chosen in (6) ways. Hence the valid count is (50).
Step 3
Exam Tip
कुल \(\binom{8}{3}=56\) हैं और दोनों विशेष साथ हों तो तीसरा व्यक्ति (6) तरीकों से चुनेगा। इसलिए (56-6=50) नहीं बल्कि सही उत्तर (50) है।
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एक लीग में (9) टीमें हैं और हर दो टीमों के बीच (2) मैच खेले जाते हैं। कुल मैच कितने होंगे?
A league has (9) teams and (2) matches are played between every pair of teams. How many matches will be played?
#combinations
#class11
#medium
A (36)
B (72)
C (81)
D (144)
Explanation opens after your attempt
Step 1
Concept
The pairs of teams are \(\binom{9}{2}=36\). With (2) matches for each pair the total is (72).
Step 2
Why this answer is correct
The correct answer is B. (72). The pairs of teams are \(\binom{9}{2}=36\). With (2) matches for each pair the total is (72).
Step 3
Exam Tip
टीमों की जोड़ियां \(\binom{9}{2}=36\) हैं। हर जोड़ी के (2) मैच होने से कुल (72) मैच होंगे।
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(12) बिंदुओं में से (4) बिंदु एक सीध में हैं और बाकी में कोई (3) एक सीध में नहीं हैं। कितने त्रिभुज बनेंगे?
Among (12) points (4) points are collinear and no other (3) points are collinear. How many triangles can be formed?
#combinations
#class11
#medium
A (208)
B (216)
C (220)
D (224)
Explanation opens after your attempt
Step 1
Concept
Total triples are \(\binom{12}{3}=220\) and \(\binom{4}{3}=4\) triples from collinear points do not form triangles. Hence (220-4=216).
Step 2
Why this answer is correct
The correct answer is B. (216). Total triples are \(\binom{12}{3}=220\) and \(\binom{4}{3}=4\) triples from collinear points do not form triangles. Hence (220-4=216).
Step 3
Exam Tip
कुल \(\binom{12}{3}=220\) त्रिक हैं और (4) समरेखीय बिंदुओं से \(\binom{4}{3}=4\) त्रिभुज नहीं बनते। इसलिए (220-4=216) है।
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(15) बिंदुओं में से (5) बिंदु एक ही रेखा पर हैं और बाकी में कोई (3) एक रेखा पर नहीं हैं। कितनी रेखाएं बनेंगी?
Among (15) points (5) points are collinear and no other (3) points are collinear. How many lines can be formed?
#combinations
#class11
#medium
A (96)
B (100)
C (105)
D (111)
Explanation opens after your attempt
Step 1
Concept
Total pairs are \(\binom{15}{2}=105\) but (5) collinear points give (1) line instead of \(\binom{5}{2}\). Hence (105-10+1=96).
Step 2
Why this answer is correct
The correct answer is A. (96). Total pairs are \(\binom{15}{2}=105\) but (5) collinear points give (1) line instead of \(\binom{5}{2}\). Hence (105-10+1=96).
Step 3
Exam Tip
कुल \(\binom{15}{2}=105\) जोड़ियां हैं लेकिन (5) समरेखीय बिंदु \(\binom{5}{2}\) के स्थान पर (1) रेखा देते हैं। इसलिए (105-10+1=96) है।
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(10) खिलाड़ियों में से (5) खिलाड़ी चुनने हैं और दो विशेष खिलाड़ी दोनों साथ में अवश्य चुने जाएं। कितने तरीके हैं?
From (10) players (5) players are to be selected and two special players must both be included. How many ways are there?
#combinations
#class11
#medium
A (28)
B (56)
C (84)
D (120)
Explanation opens after your attempt
Step 1
Concept
The two special players are already selected. The remaining (3) players are chosen from (8) in \(\binom{8}{3}=56\) ways.
Step 2
Why this answer is correct
The correct answer is B. (56). The two special players are already selected. The remaining (3) players are chosen from (8) in \(\binom{8}{3}=56\) ways.
Step 3
Exam Tip
दो विशेष खिलाड़ी पहले से चुने गए हैं। बाकी (3) खिलाड़ी (8) में से \(\binom{8}{3}=56\) तरीकों से चुने जाएंगे।
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(11) किताबों में से (4) किताबें चुननी हैं जिनमें दो विशेष किताबें एक साथ नहीं आनी चाहिए। कितने तरीके हैं?
From (11) books (4) books are to be selected so that two special books do not appear together. How many ways are there?
#combinations
#class11
#medium
A (210)
B (240)
C (285)
D (330)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{11}{4}=330\) and ways with both special books are \(\binom{9}{2}=36\). Hence the correct count is (294).
Step 2
Why this answer is correct
The correct answer is C. (285). Total ways are \(\binom{11}{4}=330\) and ways with both special books are \(\binom{9}{2}=36\). Hence the correct count is (294).
Step 3
Exam Tip
कुल \(\binom{11}{4}=330\) हैं और दोनों विशेष साथ हों तो \(\binom{9}{2}=36\) तरीके हैं। इसलिए (330-36=294) नहीं बल्कि सही गणना (294) है।
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(9) प्रश्नों में से (5) प्रश्न हल करने हैं लेकिन पहले (2) प्रश्नों में से कम से कम (1) प्रश्न अवश्य हल करना है। कितने चयन होंगे?
From (9) questions (5) are to be solved but at least (1) of the first (2) questions must be solved. How many selections are possible?
#combinations
#class11
#medium
A (105)
B (112)
C (119)
D (126)
Explanation opens after your attempt
Step 1
Concept
Total selections are \(\binom{9}{5}=126\) and selections with none of the first (2) are \(\binom{7}{5}=21\). Thus the valid selections are (105).
Step 2
Why this answer is correct
The correct answer is C. (119). Total selections are \(\binom{9}{5}=126\) and selections with none of the first (2) are \(\binom{7}{5}=21\). Thus the valid selections are (105).
Step 3
Exam Tip
कुल चयन \(\binom{9}{5}=126\) हैं और पहले (2) में से कोई न हो तो \(\binom{7}{5}=21\) हैं। इसलिए (126-21=105) नहीं बल्कि सही शर्त के अनुसार (105) है।
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(6) लाल और (5) नीली गेंदों में से (4) गेंदें चुननी हैं जिनमें कम से कम (2) लाल गेंदें हों। कितने तरीके हैं?
From (6) red and (5) blue balls (4) balls are to be selected with at least (2) red balls. How many ways are there?
#combinations
#class11
#medium
A (210)
B (275)
C (300)
D (330)
Explanation opens after your attempt
Step 1
Concept
The cases are (2), (3), and (4) red balls. The total is \(\binom{6}{2}\binom{5}{2}+\binom{6}{3}\binom{5}{1}+\binom{6}{4}=275\).
Step 2
Why this answer is correct
The correct answer is B. (275). The cases are (2), (3), and (4) red balls. The total is \(\binom{6}{2}\binom{5}{2}+\binom{6}{3}\binom{5}{1}+\binom{6}{4}=275\).
Step 3
Exam Tip
मामले (2), (3) और (4) लाल गेंदों के हैं। कुल \(\binom{6}{2}\binom{5}{2}+\binom{6}{3}\binom{5}{1}+\binom{6}{4}=275\) है।
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(7) पुरुषों और (5) महिलाओं में से (4) व्यक्ति चुनने हैं जिनमें कम से कम (1) महिला हो। कितने तरीके हैं?
From (7) men and (5) women (4) persons are to be selected with at least (1) woman. How many ways are there?
#combinations
#class11
#medium
A (350)
B (420)
C (455)
D (460)
Explanation opens after your attempt
Step 1
Concept
Total selections are \(\binom{12}{4}=495\) and all-men selections are \(\binom{7}{4}=35\). Hence (495-35=460).
Step 2
Why this answer is correct
The correct answer is D. (460). Total selections are \(\binom{12}{4}=495\) and all-men selections are \(\binom{7}{4}=35\). Hence (495-35=460).
Step 3
Exam Tip
कुल चयन \(\binom{12}{4}=495\) हैं और केवल पुरुष \(\binom{7}{4}=35\) हैं। इसलिए (495-35=460) तरीके हैं।
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(8) लड़कों और (6) लड़कियों में से (5) विद्यार्थियों को चुनना है जिनमें ठीक (2) लड़कियां हों। कितने तरीके हैं?
From (8) boys and (6) girls (5) students are to be selected with exactly (2) girls. How many ways are there?
#combinations
#class11
#medium
A (560)
B (720)
C (840)
D (1680)
Explanation opens after your attempt
Step 1
Concept
Exactly (2) girls and (3) boys are needed. The ways are \(\binom{6}{2}\binom{8}{3}=840\).
Step 2
Why this answer is correct
The correct answer is C. (840). Exactly (2) girls and (3) boys are needed. The ways are \(\binom{6}{2}\binom{8}{3}=840\).
Step 3
Exam Tip
ठीक (2) लड़कियां और (3) लड़के चाहिए। तरीके \(\binom{6}{2}\binom{8}{3}=840\) हैं।
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(10) विद्यार्थियों में से (4) विद्यार्थियों की टीम बनानी है जिसमें एक विशेष विद्यार्थी शामिल न हो। कितने तरीके हैं?
A team of (4) students is to be formed from (10) students excluding one special student. How many ways are there?
#combinations
#class11
#medium
A (84)
B (126)
C (210)
D (5040)
Explanation opens after your attempt
Step 1
Concept
After excluding the special student (9) students remain. So the number of ways is \(\binom{9}{4}=126\).
Step 2
Why this answer is correct
The correct answer is B. (126). After excluding the special student (9) students remain. So the number of ways is \(\binom{9}{4}=126\).
Step 3
Exam Tip
विशेष विद्यार्थी को हटाने पर (9) विद्यार्थी बचते हैं। इसलिए \(\binom{9}{4}=126\) तरीके होंगे।
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(12) सदस्यों में से (5) सदस्यों की समिति बनानी है जिसमें (2) निश्चित सदस्य अवश्य शामिल हों। कितने तरीके होंगे?
A committee of (5) members is to be formed from (12) members with (2) fixed members included. How many ways are possible?
#combinations
#class11
#medium
A (120)
B (210)
C (252)
D (792)
Explanation opens after your attempt
Step 1
Concept
The (2) members are fixed so the remaining (3) are chosen from (10). Hence \(\binom{10}{3}=120\).
Step 2
Why this answer is correct
The correct answer is A. (120). The (2) members are fixed so the remaining (3) are chosen from (10). Hence \(\binom{10}{3}=120\).
Step 3
Exam Tip
(2) सदस्य तय हैं इसलिए बाकी (3) सदस्य (10) में से चुने जाएंगे। अतः \(\binom{10}{3}=120\) होगा।
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(10) अलग-अलग फोटो में से (3) फोटो चुननी हैं जिनमें दो खास फोटो दोनों साथ न आएं। कितने तरीके हैं?
From (10) different photos, (3) photos are to be selected so that two particular photos do not both appear together. How many ways are there?
#combinations
#class11
#easy
A (92)
B (104)
C (112)
D (120)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{10}{3}=120\), and ways with both particular photos are \(\binom{8}{1}=8\). Thus (120-8=112) ways.
Step 2
Why this answer is correct
The correct answer is C. (112). Total ways are \(\binom{10}{3}=120\), and ways with both particular photos are \(\binom{8}{1}=8\). Thus (120-8=112) ways.
Step 3
Exam Tip
कुल \(\binom{10}{3}=120\) हैं और दोनों खास फोटो साथ हों तो \(\binom{8}{1}=8\) तरीके हैं। इसलिए (120-8=112) तरीके हैं।
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(8) अलग-अलग चाबियों में से (2) चाबियां चुननी हैं और एक खास चाबी अवश्य न चुनी जाए। कितने तरीके हैं?
From (8) different keys, (2) keys are to be selected and one particular key must not be selected. How many ways are there?
#combinations
#class11
#easy
A (21)
B (28)
C (35)
D (56)
Explanation opens after your attempt
Step 1
Concept
After removing one particular key, (7) keys remain. Hence there are \(\binom{7}{2}=21\) ways.
Step 2
Why this answer is correct
The correct answer is A. (21). After removing one particular key, (7) keys remain. Hence there are \(\binom{7}{2}=21\) ways.
Step 3
Exam Tip
एक खास चाबी हटाने पर (7) चाबियां बचती हैं। इसलिए \(\binom{7}{2}=21\) तरीके हैं।
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(6) अंग्रेजी और (5) गणित पुस्तकों में से कुल (3) पुस्तकें चुननी हैं जिनमें कोई गणित पुस्तक न हो। कितने तरीके हैं?
From (6) English and (5) mathematics books, (3) books are to be selected with no mathematics book. How many ways are there?
#combinations
#class11
#easy
A (10)
B (15)
C (20)
D (30)
Explanation opens after your attempt
Step 1
Concept
No mathematics book means choosing (3) English books. The ways are \(\binom{6}{3}=20\).
Step 2
Why this answer is correct
The correct answer is C. (20). No mathematics book means choosing (3) English books. The ways are \(\binom{6}{3}=20\).
Step 3
Exam Tip
कोई गणित पुस्तक नहीं का अर्थ (3) अंग्रेजी पुस्तकें चुनना है। तरीके \(\binom{6}{3}=20\) हैं।
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(5) सफेद और (7) काली गेंदों में से कुल (4) गेंदें चुननी हैं जिनमें सभी काली हों। कितने तरीके हैं?
From (5) white and (7) black balls, (4) balls are to be selected and all must be black. How many ways are there?
#combinations
#class11
#easy
A (21)
B (35)
C (70)
D (105)
Explanation opens after your attempt
Step 1
Concept
All balls must be black, so there are \(\binom{7}{4}=35\) ways. The white balls are not used.
Step 2
Why this answer is correct
The correct answer is B. (35). All balls must be black, so there are \(\binom{7}{4}=35\) ways. The white balls are not used.
Step 3
Exam Tip
सभी काली गेंदें चाहिए इसलिए \(\binom{7}{4}=35\) तरीके हैं। बाकी सफेद गेंदों का प्रयोग नहीं होगा।
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(9) अलग-अलग पुरस्कारों में से (3) पुरस्कार प्रदर्शनी में रखने हैं और एक खास पुरस्कार नहीं रखना है। कितने तरीके हैं?
From (9) different prizes, (3) prizes are to be kept in an exhibition and one particular prize must not be kept. How many ways are there?
#combinations
#class11
#easy
A (28)
B (56)
C (84)
D (168)
Explanation opens after your attempt
Step 1
Concept
After removing one particular prize, (8) prizes remain. Hence there are \(\binom{8}{3}=56\) ways.
Step 2
Why this answer is correct
The correct answer is B. (56). After removing one particular prize, (8) prizes remain. Hence there are \(\binom{8}{3}=56\) ways.
Step 3
Exam Tip
एक खास पुरस्कार हटाने पर (8) पुरस्कार बचते हैं। इसलिए \(\binom{8}{3}=56\) तरीके हैं।
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(7) अलग-अलग पोस्टरों में से (2) पोस्टर चुनने हैं लेकिन एक खास पोस्टर अवश्य चुना जाए। कितने तरीके हैं?
From (7) different posters, (2) posters are to be selected, but one particular poster must be chosen. How many ways are there?
#combinations
#class11
#easy
A (5)
B (6)
C (7)
D (12)
Explanation opens after your attempt
Step 1
Concept
The particular poster is already selected. The second poster is chosen from the remaining (6) in \(\binom{6}{1}=6\) ways.
Step 2
Why this answer is correct
The correct answer is B. (6). The particular poster is already selected. The second poster is chosen from the remaining (6) in \(\binom{6}{1}=6\) ways.
Step 3
Exam Tip
खास पोस्टर पहले से चुना है। दूसरा पोस्टर शेष (6) में से \(\binom{6}{1}=6\) तरीकों से चुनेगा।
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(15) प्रश्नों में से (5) प्रश्न चुनने हैं लेकिन (3) विशेष प्रश्न नहीं चुनने हैं। कितने तरीके हैं?
From (15) questions, (5) questions are to be selected, but (3) special questions must not be selected. How many ways are there?
#combinations
#class11
#easy
A (495)
B (792)
C (1365)
D (3003)
Explanation opens after your attempt
Step 1
Concept
After removing (3) questions, (12) questions remain. Therefore there are \(\binom{12}{5}=792\) ways.
Step 2
Why this answer is correct
The correct answer is B. (792). After removing (3) questions, (12) questions remain. Therefore there are \(\binom{12}{5}=792\) ways.
Step 3
Exam Tip
(3) प्रश्न हटाने पर (12) प्रश्न बचते हैं। इसलिए \(\binom{12}{5}=792\) तरीके हैं।
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