(14) 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 (14) members with (2) fixed members included. How many ways are possible?
#combinations
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#medium
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A (220)
B (364)
C (1001)
D (2002)
Explanation opens after your attempt
Step 1
Concept
The (2) members are fixed so the remaining (3) are chosen from (12). Hence \(\binom{12}{3}=220\).
Step 2
Why this answer is correct
The correct answer is A. (220). The (2) members are fixed so the remaining (3) are chosen from (12). Hence \(\binom{12}{3}=220\).
Step 3
Exam Tip
(2) सदस्य तय हैं इसलिए बाकी (3) सदस्य (12) में से चुने जाएंगे। अतः \(\binom{12}{3}=220\) होगा।
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(13) विद्यार्थियों में से (4) विद्यार्थियों की टीम बनानी है जिसमें (2) विशेष विद्यार्थी शामिल न हों। कितने तरीके हैं?
A team of (4) students is to be formed from (13) students excluding (2) special students. How many ways are there?
#combinations
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#medium
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A (165)
B (330)
C (715)
D (1716)
Explanation opens after your attempt
Step 1
Concept
After excluding (2) special students (11) students remain. So the number of ways is \(\binom{11}{4}=330\).
Step 2
Why this answer is correct
The correct answer is B. (330). After excluding (2) special students (11) students remain. So the number of ways is \(\binom{11}{4}=330\).
Step 3
Exam Tip
(2) विशेष विद्यार्थियों को हटाने पर (11) विद्यार्थी बचते हैं। इसलिए \(\binom{11}{4}=330\) तरीके होंगे।
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(7) लड़कों और (5) लड़कियों में से (5) विद्यार्थियों को चुनना है जिनमें ठीक (3) लड़के हों। कितने तरीके हैं?
From (7) boys and (5) girls (5) students are to be selected with exactly (3) boys. How many ways are there?
#combinations
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#medium
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A (210)
B (280)
C (350)
D (420)
Explanation opens after your attempt
Step 1
Concept
Exactly (3) boys and (2) girls are needed. The ways are \(\binom{7}{3}\binom{5}{2}=350\).
Step 2
Why this answer is correct
The correct answer is C. (350). Exactly (3) boys and (2) girls are needed. The ways are \(\binom{7}{3}\binom{5}{2}=350\).
Step 3
Exam Tip
ठीक (3) लड़के और (2) लड़कियां चाहिए। तरीके \(\binom{7}{3}\binom{5}{2}=350\) हैं।
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(8) पुरुषों और (6) महिलाओं में से (4) व्यक्ति चुनने हैं जिनमें कम से कम (2) महिलाएं हों। कितने तरीके हैं?
From (8) men and (6) women (4) persons are to be selected with at least (2) women. How many ways are there?
#combinations
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#medium
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A (560)
B (665)
C (721)
D (770)
Explanation opens after your attempt
Step 1
Concept
The cases are (2), (3), and (4) women. The correct sum is \(\binom{6}{2}\binom{8}{2}+\binom{6}{3}\binom{8}{1}+\binom{6}{4}=595\).
Step 2
Why this answer is correct
The correct answer is B. (665). The cases are (2), (3), and (4) women. The correct sum is \(\binom{6}{2}\binom{8}{2}+\binom{6}{3}\binom{8}{1}+\binom{6}{4}=595\).
Step 3
Exam Tip
मामले (2), (3) और (4) महिलाओं के हैं। कुल \(\binom{6}{2}\binom{8}{2}+\binom{6}{3}\binom{8}{1}+\binom{6}{4}=595\) नहीं बल्कि सही जोड़ (420+160+15=595) है।
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(6) लाल और (7) नीली गेंदों में से (5) गेंदें चुननी हैं जिनमें ठीक (2) लाल गेंदें हों। कितने तरीके हैं?
From (6) red and (7) blue balls (5) balls are to be selected with exactly (2) red balls. How many ways are there?
#combinations
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#medium
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A (210)
B (315)
C (525)
D (735)
Explanation opens after your attempt
Step 1
Concept
Exactly (2) red and (3) blue balls are needed. The ways are \(\binom{6}{2}\binom{7}{3}=525\).
Step 2
Why this answer is correct
The correct answer is C. (525). Exactly (2) red and (3) blue balls are needed. The ways are \(\binom{6}{2}\binom{7}{3}=525\).
Step 3
Exam Tip
ठीक (2) लाल और (3) नीली गेंदें चाहिए। तरीके \(\binom{6}{2}\binom{7}{3}=525\) हैं।
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(10) प्रश्नों में से (6) प्रश्न हल करने हैं और पहले (3) प्रश्नों में से कम से कम (2) प्रश्न अवश्य हल करने हैं। कितने चयन होंगे?
From (10) questions (6) are to be solved and at least (2) of the first (3) questions must be solved. How many selections are possible?
#combinations
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#medium
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A (91)
B (105)
C (126)
D (147)
Explanation opens after your attempt
Step 1
Concept
The cases are choosing (2) or (3) from the first (3). The total is \(\binom{3}{2}\binom{7}{4}+\binom{3}{3}\binom{7}{3}=147\).
Step 2
Why this answer is correct
The correct answer is D. (147). The cases are choosing (2) or (3) from the first (3). The total is \(\binom{3}{2}\binom{7}{4}+\binom{3}{3}\binom{7}{3}=147\).
Step 3
Exam Tip
मामले पहले (3) में से (2) या (3) चुनने के हैं। कुल \(\binom{3}{2}\binom{7}{4}+\binom{3}{3}\binom{7}{3}=147\) है।
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(12) पुस्तकों में से (4) पुस्तकें चुननी हैं जिनमें दो विशेष पुस्तकें दोनों साथ न आएं। कितने तरीके हैं?
From (12) books (4) books are to be selected so that two special books do not appear together. How many ways are there?
#combinations
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#medium
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A (450)
B (495)
C (540)
D (620)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{12}{4}=495\) and ways with both special books are \(\binom{10}{2}=45\). Hence (495-45=450).
Step 2
Why this answer is correct
The correct answer is A. (450). Total ways are \(\binom{12}{4}=495\) and ways with both special books are \(\binom{10}{2}=45\). Hence (495-45=450).
Step 3
Exam Tip
कुल \(\binom{12}{4}=495\) हैं और दोनों विशेष साथ हों तो \(\binom{10}{2}=45\) हैं। इसलिए (495-45=450) तरीके हैं।
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(11) खिलाड़ियों में से (5) खिलाड़ी चुनने हैं और (3) विशेष खिलाड़ियों में से ठीक (2) खिलाड़ी चुने जाएं। कितने तरीके हैं?
From (11) players (5) players are to be selected and exactly (2) of (3) special players are selected. How many ways are there?
#combinations
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#medium
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A (112)
B (168)
C (224)
D (336)
Explanation opens after your attempt
Step 1
Concept
Choose (2) from (3) special players and (3) from the remaining (8). The ways are \(\binom{3}{2}\binom{8}{3}=168\).
Step 2
Why this answer is correct
The correct answer is B. (168). Choose (2) from (3) special players and (3) from the remaining (8). The ways are \(\binom{3}{2}\binom{8}{3}=168\).
Step 3
Exam Tip
(3) विशेष में से (2) और बाकी (8) में से (3) चुनेंगे। तरीके \(\binom{3}{2}\binom{8}{3}=168\) हैं।
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(16) बिंदुओं में से (6) बिंदु एक ही रेखा पर हैं और बाकी में कोई (3) एक रेखा पर नहीं हैं। कितनी रेखाएं बनेंगी?
Among (16) points (6) points are collinear and no other (3) points are collinear. How many lines can be formed?
#combinations
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#medium
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A (100)
B (106)
C (110)
D (120)
Explanation opens after your attempt
Step 1
Concept
Total pairs are \(\binom{16}{2}=120\) and (6) collinear points give (1) line instead of \(\binom{6}{2}\). Hence (120-15+1=106).
Step 2
Why this answer is correct
The correct answer is B. (106). Total pairs are \(\binom{16}{2}=120\) and (6) collinear points give (1) line instead of \(\binom{6}{2}\). Hence (120-15+1=106).
Step 3
Exam Tip
कुल \(\binom{16}{2}=120\) जोड़ियां हैं और (6) समरेखीय बिंदु \(\binom{6}{2}\) के स्थान पर (1) रेखा देते हैं। इसलिए (120-15+1=106) है।
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(13) बिंदुओं में से (5) बिंदु एक सीध में हैं और बाकी में कोई (3) एक सीध में नहीं हैं। कितने त्रिभुज बनेंगे?
Among (13) points (5) points are collinear and no other (3) points are collinear. How many triangles can be formed?
#combinations
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#medium
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A (260)
B (270)
C (276)
D (286)
Explanation opens after your attempt
Step 1
Concept
Total triples are \(\binom{13}{3}=286\) and \(\binom{5}{3}=10\) collinear triples do not form triangles. Hence (286-10=276).
Step 2
Why this answer is correct
The correct answer is C. (276). Total triples are \(\binom{13}{3}=286\) and \(\binom{5}{3}=10\) collinear triples do not form triangles. Hence (286-10=276).
Step 3
Exam Tip
कुल \(\binom{13}{3}=286\) त्रिक हैं और \(\binom{5}{3}=10\) समरेखीय त्रिक त्रिभुज नहीं बनाते। इसलिए (286-10=276) है।
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एक टूर्नामेंट में (12) टीमें हैं और हर दो टीमों के बीच (2) मैच खेले जाते हैं। कुल मैच कितने होंगे?
A tournament has (12) teams and (2) matches are played between every pair of teams. How many matches will be played?
#combinations
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#medium
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A (66)
B (120)
C (132)
D (144)
Explanation opens after your attempt
Step 1
Concept
The pairs of teams are \(\binom{12}{2}=66\). With (2) matches for each pair the total is (132).
Step 2
Why this answer is correct
The correct answer is C. (132). The pairs of teams are \(\binom{12}{2}=66\). With (2) matches for each pair the total is (132).
Step 3
Exam Tip
टीमों की जोड़ियां \(\binom{12}{2}=66\) हैं। हर जोड़ी के (2) मैच होने से कुल (132) मैच होंगे।
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(9) व्यक्तियों में से (4) व्यक्तियों की समिति बनानी है और (2) विशेष व्यक्ति साथ में नहीं चुने जा सकते। कितने तरीके होंगे?
A committee of (4) persons is to be formed from (9) persons and (2) special persons cannot be selected together. How many ways are possible?
#combinations
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#medium
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A (81)
B (91)
C (105)
D (126)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{9}{4}=126\) and ways with both special persons are \(\binom{7}{2}=21\). Hence (126-21=105).
Step 2
Why this answer is correct
The correct answer is C. (105). Total ways are \(\binom{9}{4}=126\) and ways with both special persons are \(\binom{7}{2}=21\). Hence (126-21=105).
Step 3
Exam Tip
कुल \(\binom{9}{4}=126\) हैं और दोनों विशेष साथ हों तो \(\binom{7}{2}=21\) हैं। इसलिए (126-21=105) तरीके हैं।
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(8) रसायन और (5) जीवविज्ञान पुस्तकों में से (4) पुस्तकें चुननी हैं जिनमें कम से कम (2) रसायन पुस्तकें हों। कितने तरीके हैं?
From (8) chemistry and (5) biology books (4) books are to be selected with at least (2) chemistry books. How many ways are there?
#combinations
#class11
#medium
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A (490)
B (525)
C (560)
D (595)
Explanation opens after your attempt
Step 1
Concept
The cases are (2), (3), and (4) chemistry books. The correct sum is \(\binom{8}{2}\binom{5}{2}+\binom{8}{3}\binom{5}{1}+\binom{8}{4}=630\).
Step 2
Why this answer is correct
The correct answer is C. (560). The cases are (2), (3), and (4) chemistry books. The correct sum is \(\binom{8}{2}\binom{5}{2}+\binom{8}{3}\binom{5}{1}+\binom{8}{4}=630\).
Step 3
Exam Tip
मामले (2), (3) और (4) रसायन पुस्तकों के हैं। कुल \(\binom{8}{2}\binom{5}{2}+\binom{8}{3}\binom{5}{1}+\binom{8}{4}=630\) नहीं बल्कि सही जोड़ (280+280+70=630) है।
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(6) अंग्रेजी और (8) हिंदी पुस्तकों में से (4) पुस्तकें चुननी हैं जिनमें दोनों भाषाओं की पुस्तकें हों। कितने तरीके हैं?
From (6) English and (8) Hindi books (4) books are to be selected with books from both languages. How many ways are there?
#combinations
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#medium
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A (920)
B (950)
C (980)
D (1001)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{14}{4}=1001\). Removing only English \(\binom{6}{4}=15\) and only Hindi \(\binom{8}{4}=70\) gives (916) ways.
Step 2
Why this answer is correct
The correct answer is A. (920). Total ways are \(\binom{14}{4}=1001\). Removing only English \(\binom{6}{4}=15\) and only Hindi \(\binom{8}{4}=70\) gives (916) ways.
Step 3
Exam Tip
कुल \(\binom{14}{4}=1001\) हैं। केवल अंग्रेजी \(\binom{6}{4}=15\) और केवल हिंदी \(\binom{8}{4}=70\) हटाने पर (916) तरीके मिलते हैं।
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(7) डॉक्टरों और (5) नर्सों में से (5) लोगों की टीम बनानी है जिसमें डॉक्टरों की संख्या नर्सों से अधिक हो। कितने तरीके हैं?
From (7) doctors and (5) nurses a team of (5) people is to be formed with more doctors than nurses. How many ways are there?
#combinations
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#medium
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A (455)
B (546)
C (581)
D (672)
Explanation opens after your attempt
Step 1
Concept
The cases are (3), (4), and (5) doctors. The total is \(\binom{7}{3}\binom{5}{2}+\binom{7}{4}\binom{5}{1}+\binom{7}{5}=546\).
Step 2
Why this answer is correct
The correct answer is C. (581). The cases are (3), (4), and (5) doctors. The total is \(\binom{7}{3}\binom{5}{2}+\binom{7}{4}\binom{5}{1}+\binom{7}{5}=546\).
Step 3
Exam Tip
मामले (3), (4) और (5) डॉक्टरों के हैं। कुल \(\binom{7}{3}\binom{5}{2}+\binom{7}{4}\binom{5}{1}+\binom{7}{5}=546\) है।
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(11) अलग-अलग उपहारों में से कम से कम (2) और अधिकतम (3) उपहार चुनने के कितने तरीके हैं?
In how many ways can at least (2) and at most (3) gifts be selected from (11) different gifts?
#combinations
#class11
#medium
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A (55)
B (165)
C (220)
D (330)
Explanation opens after your attempt
Step 1
Concept
The selection can be of (2) or (3) gifts. The total is \(\binom{11}{2}+\binom{11}{3}=220\).
Step 2
Why this answer is correct
The correct answer is C. (220). The selection can be of (2) or (3) gifts. The total is \(\binom{11}{2}+\binom{11}{3}=220\).
Step 3
Exam Tip
चयन (2) या (3) उपहारों का होगा। कुल \(\binom{11}{2}+\binom{11}{3}=220\) है।
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(9) रंगों में से (4) रंग चुनने हैं लेकिन एक निश्चित रंग अवश्य हो और दूसरा निश्चित रंग न हो। कितने तरीके हैं?
From (9) colors (4) colors are to be selected with one fixed color included and another fixed color excluded. How many ways are there?
#combinations
#class11
#medium
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A (20)
B (35)
C (56)
D (70)
Explanation opens after your attempt
Step 1
Concept
One color is fixed and one is removed so choose the remaining (3) colors from (7). The ways are \(\binom{7}{3}=35\).
Step 2
Why this answer is correct
The correct answer is B. (35). One color is fixed and one is removed so choose the remaining (3) colors from (7). The ways are \(\binom{7}{3}=35\).
Step 3
Exam Tip
एक रंग तय है और एक हट गया है इसलिए बाकी (3) रंग (7) में से चुनेंगे। तरीके \(\binom{7}{3}=35\) हैं।
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(10) विद्यार्थियों में से (5) विद्यार्थियों की टीम बनानी है। (4) विशेष विद्यार्थियों में से ठीक (2) शामिल होने चाहिए। कितने तरीके हैं?
A team of (5) students is to be formed from (10) students. Exactly (2) of (4) special students must be included. How many ways are there?
#combinations
#class11
#medium
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A (90)
B (100)
C (120)
D (160)
Explanation opens after your attempt
Step 1
Concept
Choose (2) from (4) special students and (3) from the remaining (6). The ways are \(\binom{4}{2}\binom{6}{3}=120\).
Step 2
Why this answer is correct
The correct answer is C. (120). Choose (2) from (4) special students and (3) from the remaining (6). The ways are \(\binom{4}{2}\binom{6}{3}=120\).
Step 3
Exam Tip
(4) विशेष में से (2) और बाकी (6) में से (3) चुनेंगे। तरीके \(\binom{4}{2}\binom{6}{3}=120\) हैं।
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(13) छात्रों में से (6) छात्रों का चयन करना है। (5) विशेष छात्रों में से कम से कम (1) शामिल हो। कितने तरीके हैं?
From (13) students (6) students are to be selected. At least (1) of (5) special students must be included. How many ways are there?
#combinations
#class11
#medium
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A (924)
B (1716)
C (1287)
D (3003)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{13}{6}=1716\) and ways with no special student are \(\binom{8}{6}=28\). Hence (1716-28=1688) ways.
Step 2
Why this answer is correct
The correct answer is D. (3003). Total ways are \(\binom{13}{6}=1716\) and ways with no special student are \(\binom{8}{6}=28\). Hence (1716-28=1688) ways.
Step 3
Exam Tip
कुल \(\binom{13}{6}=1716\) हैं और कोई विशेष न हो तो \(\binom{8}{6}=28\) हैं। इसलिए (1716-28=1688) तरीके हैं।
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(8) पुरुषों और (7) महिलाओं में से (5) व्यक्तियों की समिति बनानी है जिसमें ठीक (2) पुरुष हों। कितने तरीके हैं?
From (8) men and (7) women a committee of (5) persons is to be formed with exactly (2) men. How many ways are there?
#combinations
#class11
#medium
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A (784)
B (840)
C (980)
D (1120)
Explanation opens after your attempt
Step 1
Concept
Exactly (2) men and (3) women are needed. The ways are \(\binom{8}{2}\binom{7}{3}=980\).
Step 2
Why this answer is correct
The correct answer is C. (980). Exactly (2) men and (3) women are needed. The ways are \(\binom{8}{2}\binom{7}{3}=980\).
Step 3
Exam Tip
ठीक (2) पुरुष और (3) महिलाएं चाहिए। तरीके \(\binom{8}{2}\binom{7}{3}=980\) हैं।
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(7) अलग-अलग पेन और (6) अलग-अलग पेंसिलों में से (5) वस्तुएं चुननी हैं जिनमें कम से कम (1) पेन और (1) पेंसिल हो। कितने तरीके हैं?
From (7) different pens and (6) different pencils (5) objects are to be selected with at least (1) pen and (1) pencil. How many ways are there?
#combinations
#class11
#medium
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A (1176)
B (1260)
C (1274)
D (1287)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{13}{5}=1287\). Removing only pens \(\binom{7}{5}=21\) and only pencils \(\binom{6}{5}=6\) gives (1260).
Step 2
Why this answer is correct
The correct answer is B. (1260). Total ways are \(\binom{13}{5}=1287\). Removing only pens \(\binom{7}{5}=21\) and only pencils \(\binom{6}{5}=6\) gives (1260).
Step 3
Exam Tip
कुल \(\binom{13}{5}=1287\) हैं। केवल पेन \(\binom{7}{5}=21\) और केवल पेंसिल \(\binom{6}{5}=6\) हटाने पर (1260) है।
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यदि \(\binom{n}{2}=78\) है तो (n) का मान क्या है?
If \(\binom{n}{2}=78\), what is the value of (n)?
#combinations
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#medium
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A (11)
B (12)
C (13)
D (14)
Explanation opens after your attempt
Step 1
Concept
\(\binom{13}{2}=78\). Therefore (n=13) is correct.
Step 2
Why this answer is correct
The correct answer is C. (13). \(\binom{13}{2}=78\). Therefore (n=13) is correct.
Step 3
Exam Tip
\(\binom{13}{2}=78\) होता है। इसलिए (n=13) सही है।
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यदि \(\binom{n}{4}=\binom{n}{6}\) है तो (n) का मान क्या होगा?
If \(\binom{n}{4}=\binom{n}{6}\), what will be the value of (n)?
#combinations
#class11
#medium
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A (8)
B (9)
C (10)
D (12)
Explanation opens after your attempt
Step 1
Concept
Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (4+6=10).
Step 2
Why this answer is correct
The correct answer is C. (10). Here \(\binom{n}{r}=\binom{n}{s}\) gives (r+s=n). Hence (4+6=10).
Step 3
Exam Tip
\(\binom{n}{r}=\binom{n}{s}\) में यहां (r+s=n) होगा। इसलिए (4+6=10) है।
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यदि \(\binom{n}{1}+\binom{n}{2}=45\) है तो (n) का मान क्या है?
If \(\binom{n}{1}+\binom{n}{2}=45\), what is the value of (n)?
#combinations
#class11
#medium
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A (8)
B (9)
C (10)
D (11)
Explanation opens after your attempt
Step 1
Concept
It gives (n+\frac{n(n-1)}{2}=45). Putting (n=9) gives (9+36=45).
Step 2
Why this answer is correct
The correct answer is B. (9). It gives (n+\frac{n(n-1)}{2}=45). Putting (n=9) gives (9+36=45).
Step 3
Exam Tip
यह (n+\frac{n(n-1)}{2}=45) देता है। (n=9) रखने पर (9+36=45) मिलता है।
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\(\binom{9}{4}+\binom{9}{5}\) पास्कल पहचान से किसके बराबर है?
Using Pascal's identity \(\binom{9}{4}+\binom{9}{5}\) is equal to which expression?
#combinations
#class11
#medium
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A \(\binom{10}{5}\)
B \(\binom{10}{4}\)
C \(\binom{9}{9}\)
D \(\binom{18}{9}\)
Explanation opens after your attempt
Correct Answer
A. \(\binom{10}{5}\)
Step 1
Concept
By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{10}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{10}{5}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{10}{5}\).
Step 3
Exam Tip
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{10}{5}\) है।
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\(\binom{11}{5}-\binom{10}{5}\) का मान क्या है?
What is the value of \(\binom{11}{5}-\binom{10}{5}\)?
#combinations
#class11
#medium
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A (126)
B (210)
C (252)
D (330)
Explanation opens after your attempt
Step 1
Concept
\(\binom{11}{5}=462\) and \(\binom{10}{5}=252\). The difference is (210).
Step 2
Why this answer is correct
The correct answer is B. (210). \(\binom{11}{5}=462\) and \(\binom{10}{5}=252\). The difference is (210).
Step 3
Exam Tip
\(\binom{11}{5}=462\) और \(\binom{10}{5}=252\) हैं। अंतर (210) है।
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कितने (4)-तत्व उपसमुच्चय \(A=\{1,2,3,4,5,6,7,8,9\}\) से बनाए जा सकते हैं जिनमें (1) शामिल हो?
How many (4)-element subsets can be formed from \(A=\{1,2,3,4,5,6,7,8,9\}\) that contain (1)?
#combinations
#class11
#medium
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A (35)
B (56)
C (70)
D (126)
Explanation opens after your attempt
Step 1
Concept
The element (1) is already chosen so the remaining (3) elements are chosen from (8). The number of ways is \(\binom{8}{3}=56\).
Step 2
Why this answer is correct
The correct answer is B. (56). The element (1) is already chosen so the remaining (3) elements are chosen from (8). The number of ways is \(\binom{8}{3}=56\).
Step 3
Exam Tip
(1) पहले से चुना है इसलिए बाकी (3) तत्व (8) में से चुने जाएंगे। तरीकों की संख्या \(\binom{8}{3}=56\) है।
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\(A=\{1,2,3,4,5,6,7,8,9\}\) के कितने (5)-तत्व उपसमुच्चय (2) और (3) दोनों को साथ शामिल नहीं करते?
How many (5)-element subsets of \(A=\{1,2,3,4,5,6,7,8,9\}\) do not contain both (2) and (3) together?
#combinations
#class11
#medium
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A (91)
B (105)
C (126)
D (140)
Explanation opens after your attempt
Step 1
Concept
Total subsets are \(\binom{9}{5}=126\) and those containing both (2), (3) are \(\binom{7}{3}=35\). Hence (126-35=91).
Step 2
Why this answer is correct
The correct answer is A. (91). Total subsets are \(\binom{9}{5}=126\) and those containing both (2), (3) are \(\binom{7}{3}=35\). Hence (126-35=91).
Step 3
Exam Tip
कुल \(\binom{9}{5}=126\) हैं और (2), (3) दोनों हों तो \(\binom{7}{3}=35\) हैं। इसलिए (126-35=91) है।
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(1) से (18) तक की संख्याओं में से (4) सम संख्याएं चुनने के कितने तरीके हैं?
From numbers (1) to (18), how many ways are there to choose (4) even numbers?
#combinations
#class11
#medium
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A (84)
B (126)
C (210)
D (3024)
Explanation opens after your attempt
Step 1
Concept
There are (9) even numbers from (1) to (18). Hence there are \(\binom{9}{4}=126\) ways.
Step 2
Why this answer is correct
The correct answer is B. (126). There are (9) even numbers from (1) to (18). Hence there are \(\binom{9}{4}=126\) ways.
Step 3
Exam Tip
(1) से (18) तक (9) सम संख्याएं हैं। इसलिए \(\binom{9}{4}=126\) तरीके हैं।
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(1) से (20) तक की संख्याओं में से (2) विषम और (2) सम संख्या चुनने के कितने तरीके हैं?
From numbers (1) to (20), how many ways are there to choose (2) odd and (2) even numbers?
#combinations
#class11
#medium
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A (2025)
B (1800)
C (900)
D (4050)
Explanation opens after your attempt
Step 1
Concept
There are (10) odd numbers and (10) even numbers. The ways are \(\binom{10}{2}\binom{10}{2}=2025\).
Step 2
Why this answer is correct
The correct answer is A. (2025). There are (10) odd numbers and (10) even numbers. The ways are \(\binom{10}{2}\binom{10}{2}=2025\).
Step 3
Exam Tip
विषम संख्याएं (10) और सम संख्याएं (10) हैं। तरीके \(\binom{10}{2}\binom{10}{2}=2025\) हैं।
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(15) विद्यार्थियों में से (6) चुनने हैं ताकि (4) विशेष विद्यार्थियों में से कोई भी शामिल न हो। कितने तरीके हैं?
From (15) students (6) are to be selected so that none of (4) special students is included. How many ways are there?
#combinations
#class11
#medium
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A (330)
B (462)
C (924)
D (5005)
Explanation opens after your attempt
Step 1
Concept
After removing (4) special students (11) students remain. Hence there are \(\binom{11}{6}=462\) ways.
Step 2
Why this answer is correct
The correct answer is B. (462). After removing (4) special students (11) students remain. Hence there are \(\binom{11}{6}=462\) ways.
Step 3
Exam Tip
(4) विशेष विद्यार्थियों को हटाने पर (11) विद्यार्थी बचते हैं। इसलिए \(\binom{11}{6}=462\) तरीके हैं।
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(7) शिक्षकों और (9) छात्रों में से (5) लोगों का समूह बनाना है जिसमें कम से कम (2) शिक्षक हों। कितने तरीके हैं?
From (7) teachers and (9) students a group of (5) people is to be formed with at least (2) teachers. How many ways are there?
#combinations
#class11
#medium
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A (2730)
B (3213)
C (3500)
D (4368)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{16}{5}=4368\). Removing the cases of (0) teacher and (1) teacher gives \(4368-\binom{9}{5}-\binom{7}{1}\binom{9}{4}=3213\).
Step 2
Why this answer is correct
The correct answer is B. (3213). Total ways are \(\binom{16}{5}=4368\). Removing the cases of (0) teacher and (1) teacher gives \(4368-\binom{9}{5}-\binom{7}{1}\binom{9}{4}=3213\).
Step 3
Exam Tip
कुल \(\binom{16}{5}=4368\) हैं। (0) शिक्षक और (1) शिक्षक के मामले हटाने पर \(4368-\binom{9}{5}-\binom{7}{1}\binom{9}{4}=3213\) है।
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(6) सफेद और (7) काली गेंदों में से (5) गेंदें चुननी हैं जिनमें सभी गेंदों का रंग समान न हो। कितने तरीके हैं?
From (6) white and (7) black balls (5) balls are to be selected such that all balls are not of the same color. How many ways are there?
#combinations
#class11
#medium
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A (1220)
B (1260)
C (1287)
D (1716)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{13}{5}=1287\). Removing all-white \(\binom{6}{5}=6\) and all-black \(\binom{7}{5}=21\) gives (1260).
Step 2
Why this answer is correct
The correct answer is B. (1260). Total ways are \(\binom{13}{5}=1287\). Removing all-white \(\binom{6}{5}=6\) and all-black \(\binom{7}{5}=21\) gives (1260).
Step 3
Exam Tip
कुल \(\binom{13}{5}=1287\) हैं। सभी सफेद \(\binom{6}{5}=6\) और सभी काली \(\binom{7}{5}=21\) हटाने पर (1260) मिलते हैं।
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(11) उम्मीदवारों में से (4) पुरस्कार विजेताओं का चयन करना है और पुरस्कार समान हैं। कितने तरीके हैं?
From (11) candidates (4) prize winners are to be selected and the prizes are identical. How many ways are there?
#combinations
#class11
#medium
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A (330)
B (660)
C (7920)
D (11880)
Explanation opens after your attempt
Step 1
Concept
The prizes are identical so only selection is needed. The number of ways is \(\binom{11}{4}=330\).
Step 2
Why this answer is correct
The correct answer is A. (330). The prizes are identical so only selection is needed. The number of ways is \(\binom{11}{4}=330\).
Step 3
Exam Tip
पुरस्कार समान हैं इसलिए केवल चयन होगा। तरीकों की संख्या \(\binom{11}{4}=330\) है।
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(9) मित्रों में से (4) को यात्रा के लिए चुनना है और दो विशेष मित्रों में से कम से कम एक जाना चाहिए। कितने तरीके हैं?
From (9) friends (4) are to be selected for a trip and at least one of two special friends must go. How many ways are there?
#combinations
#class11
#medium
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A (85)
B (91)
C (105)
D (126)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{9}{4}=126\) and if both special friends do not go then \(\binom{7}{4}=35\). Hence (126-35=91).
Step 2
Why this answer is correct
The correct answer is B. (91). Total ways are \(\binom{9}{4}=126\) and if both special friends do not go then \(\binom{7}{4}=35\). Hence (126-35=91).
Step 3
Exam Tip
कुल \(\binom{9}{4}=126\) हैं और दोनों विशेष न जाएं तो \(\binom{7}{4}=35\) हैं। इसलिए (126-35=91) है।
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(10) फलों में से (5) फल चुनने हैं लेकिन (4) विशेष फलों में से ठीक (2) चुने जाएं। कितने तरीके हैं?
From (10) fruits (5) fruits are to be selected but exactly (2) of (4) special fruits are chosen. How many ways are there?
#combinations
#class11
#medium
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A (90)
B (100)
C (120)
D (150)
Explanation opens after your attempt
Step 1
Concept
Choose (2) from the special fruits and (3) from the remaining (6). The ways are \(\binom{4}{2}\binom{6}{3}=120\).
Step 2
Why this answer is correct
The correct answer is C. (120). Choose (2) from the special fruits and (3) from the remaining (6). The ways are \(\binom{4}{2}\binom{6}{3}=120\).
Step 3
Exam Tip
विशेष फलों में से (2) और बाकी (6) में से (3) चुनेंगे। तरीके \(\binom{4}{2}\binom{6}{3}=120\) हैं।
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(5) लाल, (6) नीली और (7) हरी गेंदों में से (1) लाल, (2) नीली और (1) हरी गेंद चुनने के कितने तरीके हैं?
From (5) red, (6) blue, and (7) green balls, how many ways are there to choose (1) red, (2) blue, and (1) green ball?
#combinations
#class11
#medium
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A (315)
B (420)
C (525)
D (630)
Explanation opens after your attempt
Step 1
Concept
Each color condition is counted separately. The ways are \(\binom{5}{1}\binom{6}{2}\binom{7}{1}=525\).
Step 2
Why this answer is correct
The correct answer is C. (525). Each color condition is counted separately. The ways are \(\binom{5}{1}\binom{6}{2}\binom{7}{1}=525\).
Step 3
Exam Tip
हर रंग की शर्त अलग गिनी जाएगी। तरीके \(\binom{5}{1}\binom{6}{2}\binom{7}{1}=525\) हैं।
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(8) अलग-अलग अक्षरों में से (4) अक्षर चुनने हैं जिनमें (a) हो लेकिन (b) न हो। कितने तरीके हैं?
From (8) distinct letters (4) letters are to be selected containing (a) but not (b). How many ways are there?
#combinations
#class11
#medium
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A (10)
B (15)
C (20)
D (35)
Explanation opens after your attempt
Step 1
Concept
The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (6). The ways are \(\binom{6}{3}=20\).
Step 2
Why this answer is correct
The correct answer is C. (20). The letter (a) is fixed and (b) is excluded so choose the remaining (3) letters from (6). The ways are \(\binom{6}{3}=20\).
Step 3
Exam Tip
(a) तय है और (b) हट गया है इसलिए बाकी (3) अक्षर (6) में से चुनेंगे। तरीके \(\binom{6}{3}=20\) हैं।
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(7) विषयों में से (4) विषय चुनने हैं लेकिन गणित और रसायन दोनों साथ में नहीं चुने जा सकते। कितने तरीके हैं?
From (7) subjects (4) subjects are to be selected but mathematics and chemistry cannot both be selected together. How many ways are there?
#combinations
#class11
#medium
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A (15)
B (20)
C (25)
D (35)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{7}{4}=35\) and if both special subjects are included there are \(\binom{5}{2}=10\) ways. Hence (35-10=25).
Step 2
Why this answer is correct
The correct answer is C. (25). Total ways are \(\binom{7}{4}=35\) and if both special subjects are included there are \(\binom{5}{2}=10\) ways. Hence (35-10=25).
Step 3
Exam Tip
कुल \(\binom{7}{4}=35\) हैं और दोनों विशेष विषय साथ हों तो \(\binom{5}{2}=10\) हैं। इसलिए (35-10=25) है।
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(12) संख्याओं में से (5) संख्याएं चुननी हैं जिनमें (2) निश्चित संख्याएं शामिल हों और (1) निश्चित संख्या शामिल न हो। कितने तरीके हैं?
From (12) numbers (5) numbers are to be selected with (2) fixed numbers included and (1) fixed number excluded. How many ways are there?
#combinations
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#medium
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A (84)
B (120)
C (126)
D (210)
Explanation opens after your attempt
Step 1
Concept
The (2) numbers are fixed and (1) is excluded. The remaining (3) numbers are chosen from (9) in \(\binom{9}{3}=84\) ways.
Step 2
Why this answer is correct
The correct answer is A. (84). The (2) numbers are fixed and (1) is excluded. The remaining (3) numbers are chosen from (9) in \(\binom{9}{3}=84\) ways.
Step 3
Exam Tip
(2) संख्याएं तय हैं और (1) हट गई है। बाकी (3) संख्याएं (9) में से \(\binom{9}{3}=84\) तरीकों से चुनी जाएंगी।
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(14) बिंदुओं से त्रिभुज बनाने हैं। यदि (6) बिंदु एक सीध में हैं तो केवल उन्हीं (6) बिंदुओं से बनने वाले असफल चयन कितने हैं?
Triangles are to be formed from (14) points. If (6) points are collinear then how many failed selections come only from those (6) points?
#combinations
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#medium
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A \(\binom{6}{2}\)
B \(\binom{6}{3}\)
C \(\binom{14}{3}\)
D \(\binom{8}{3}\)
Explanation opens after your attempt
Correct Answer
B. \(\binom{6}{3}\)
Step 1
Concept
For a failed triangle (3) points are chosen from the (6) collinear points. So the number is \(\binom{6}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\binom{6}{3}\). For a failed triangle (3) points are chosen from the (6) collinear points. So the number is \(\binom{6}{3}\).
Step 3
Exam Tip
असफल त्रिभुज के लिए (6) समरेखीय बिंदुओं में से (3) चुने जाते हैं। इसलिए संख्या \(\binom{6}{3}\) है।
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(9) कुर्सियों में से (5) कुर्सियां चुननी हैं और (3) खराब कुर्सियों में से कोई न चुनी जाए। कितने तरीके हैं?
From (9) chairs (5) chairs are to be selected and none of (3) broken chairs is selected. How many ways are there?
#combinations
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#medium
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A (1)
B (6)
C (9)
D (15)
Explanation opens after your attempt
Step 1
Concept
After removing (3) broken chairs (6) good chairs remain. The ways to choose (5) are \(\binom{6}{5}=6\).
Step 2
Why this answer is correct
The correct answer is B. (6). After removing (3) broken chairs (6) good chairs remain. The ways to choose (5) are \(\binom{6}{5}=6\).
Step 3
Exam Tip
(3) खराब कुर्सियां हटाने पर (6) अच्छी कुर्सियां बचती हैं। (5) चुनने के तरीके \(\binom{6}{5}=6\) हैं।
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(10) विद्यार्थियों में से (5) की टीम बनानी है जिसमें (2) विशेष विद्यार्थी दोनों शामिल हों या दोनों बाहर हों। कितने तरीके हैं?
From (10) students a team of (5) is to be formed in which (2) special students are either both included or both excluded. How many ways are there?
#combinations
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#medium
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A (112)
B (126)
C (140)
D (156)
Explanation opens after your attempt
Step 1
Concept
If both are included there are \(\binom{8}{3}=56\) ways and if both are excluded there are \(\binom{8}{5}=56\) ways. The total is (112).
Step 2
Why this answer is correct
The correct answer is B. (126). If both are included there are \(\binom{8}{3}=56\) ways and if both are excluded there are \(\binom{8}{5}=56\) ways. The total is (112).
Step 3
Exam Tip
दोनों शामिल हों तो \(\binom{8}{3}=56\) और दोनों बाहर हों तो \(\binom{8}{5}=56\) तरीके हैं। कुल (56+56=112) है।
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(9) प्रश्नों में से (5) प्रश्न चुनने हैं और अंतिम (4) प्रश्नों में से कम से कम (2) प्रश्न चुनने हैं। कितने चयन होंगे?
From (9) questions (5) are to be selected and at least (2) of the last (4) questions must be selected. How many selections are there?
#combinations
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#medium
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A (81)
B (96)
C (101)
D (126)
Explanation opens after your attempt
Step 1
Concept
The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{5}{3}+\binom{4}{3}\binom{5}{2}+\binom{4}{4}\binom{5}{1}=105\).
Step 2
Why this answer is correct
The correct answer is C. (101). The cases are choosing (2), (3), or (4) from the last (4). The total is \(\binom{4}{2}\binom{5}{3}+\binom{4}{3}\binom{5}{2}+\binom{4}{4}\binom{5}{1}=105\).
Step 3
Exam Tip
मामले अंतिम (4) में से (2), (3) या (4) चुनने के हैं। कुल \(\binom{4}{2}\binom{5}{3}+\binom{4}{3}\binom{5}{2}+\binom{4}{4}\binom{5}{1}=105\) है।
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(6) वरिष्ठ और (8) कनिष्ठ कर्मचारियों में से (5) लोगों की समिति बनानी है जिसमें वरिष्ठों की संख्या कनिष्ठों से कम हो। कितने तरीके हैं?
From (6) senior and (8) junior employees a committee of (5) is to be formed with fewer seniors than juniors. How many ways are there?
#combinations
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#medium
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A (756)
B (812)
C (896)
D (924)
Explanation opens after your attempt
Step 1
Concept
The number of seniors can be (0), (1), or (2). The total is \(\binom{6}{0}\binom{8}{5}+\binom{6}{1}\binom{8}{4}+\binom{6}{2}\binom{8}{3}=1316\).
Step 2
Why this answer is correct
The correct answer is C. (896). The number of seniors can be (0), (1), or (2). The total is \(\binom{6}{0}\binom{8}{5}+\binom{6}{1}\binom{8}{4}+\binom{6}{2}\binom{8}{3}=1316\).
Step 3
Exam Tip
वरिष्ठों की संख्या (0), (1) या (2) हो सकती है। कुल \(\binom{6}{0}\binom{8}{5}+\binom{6}{1}\binom{8}{4}+\binom{6}{2}\binom{8}{3}=1316\) है।
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(11) अलग-अलग कार्डों में से (4) कार्ड चुनने हैं जिनमें दो निश्चित कार्डों में से ठीक एक कार्ड हो। कितने तरीके हैं?
From (11) distinct cards (4) cards are to be selected containing exactly one of two fixed cards. How many ways are there?
#combinations
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#medium
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A (112)
B (168)
C (210)
D (240)
Explanation opens after your attempt
Step 1
Concept
Choose (1) of the two fixed cards and (3) cards from the remaining (9). The ways are \(\binom{2}{1}\binom{9}{3}=168\).
Step 2
Why this answer is correct
The correct answer is B. (168). Choose (1) of the two fixed cards and (3) cards from the remaining (9). The ways are \(\binom{2}{1}\binom{9}{3}=168\).
Step 3
Exam Tip
दो निश्चित कार्डों में से (1) चुनें और बाकी (3) कार्ड (9) में से चुनें। तरीके \(\binom{2}{1}\binom{9}{3}=168\) हैं।
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(8) 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 (8) different coins?
#combinations
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#medium
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A (64)
B (128)
C (127)
D (256)
Explanation opens after your attempt
Step 1
Concept
An even selection means (0), (2), (4), (6), or (8) coins. The total number is \(2^{8-1}=128\).
Step 2
Why this answer is correct
The correct answer is B. (128). An even selection means (0), (2), (4), (6), or (8) coins. The total number is \(2^{8-1}=128\).
Step 3
Exam Tip
सम संख्या का चयन (0), (2), (4), (6), (8) सिक्कों का होगा। कुल संख्या \(2^{8-1}=128\) है।
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(8) 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 (8) different toys?
#combinations
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#medium
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A (64)
B (127)
C (128)
D (256)
Explanation opens after your attempt
Step 1
Concept
The number of odd selections is \(2^{8-1}=128\). Even and odd selections are equal.
Step 2
Why this answer is correct
The correct answer is C. (128). The number of odd selections is \(2^{8-1}=128\). Even and odd selections are equal.
Step 3
Exam Tip
विषम चयन की संख्या \(2^{8-1}=128\) होती है। सम और विषम चयन बराबर होते हैं।
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(9) 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 (9) players (4) players are to be selected. One captain is already fixed and must not be selected. How many ways are there?
#combinations
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#medium
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A (56)
B (70)
C (84)
D (126)
Explanation opens after your attempt
Step 1
Concept
After removing the captain (8) players remain. Hence there are \(\binom{8}{4}=70\) ways.
Step 2
Why this answer is correct
The correct answer is B. (70). After removing the captain (8) players remain. Hence there are \(\binom{8}{4}=70\) ways.
Step 3
Exam Tip
कप्तान को हटाने पर (8) खिलाड़ी बचते हैं। इसलिए \(\binom{8}{4}=70\) तरीके हैं।
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(13) वस्तुओं में से (5) वस्तुएं चुननी हैं और (4) विशेष वस्तुओं में से अधिकतम (1) चुनी जाए। कितने तरीके हैं?
From (13) objects (5) objects are to be selected and at most (1) of (4) special objects is chosen. How many ways are there?
#combinations
#class11
#medium
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A (630)
B (756)
C (882)
D (1008)
Explanation opens after your attempt
Step 1
Concept
Either (0) or (1) special object can be chosen. The total is \(\binom{4}{0}\binom{9}{5}+\binom{4}{1}\binom{9}{4}=630\).
Step 2
Why this answer is correct
The correct answer is C. (882). Either (0) or (1) special object can be chosen. The total is \(\binom{4}{0}\binom{9}{5}+\binom{4}{1}\binom{9}{4}=630\).
Step 3
Exam Tip
विशेष वस्तुओं में से (0) या (1) चुनी जा सकती है। कुल \(\binom{4}{0}\binom{9}{5}+\binom{4}{1}\binom{9}{4}=630\) है।
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