(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
#class11
#medium
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।
Login to save your score, XP, coins and progress. Login
(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
#class11
#medium
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।
Login to save your score, XP, coins and progress. Login
(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
#class11
#medium
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।
Login to save your score, XP, coins and progress. Login
(12) वस्तुओं में से (4) वस्तुएं चुननी हैं और (3) विशेष वस्तुओं में से अधिकतम (1) चुनी जाए। कितने तरीके हैं?
From (12) objects (4) objects are to be selected and at most (1) of (3) special objects is chosen. How many ways are there?
#combinations
#class11
#medium
A (336)
B (360)
C (405)
D (450)
Explanation opens after your attempt
Step 1
Concept
Either (0) or (1) special object can be chosen. The total is \(\binom{3}{0}\binom{9}{4}+\binom{3}{1}\binom{9}{3}=378\).
Step 2
Why this answer is correct
The correct answer is A. (336). Either (0) or (1) special object can be chosen. The total is \(\binom{3}{0}\binom{9}{4}+\binom{3}{1}\binom{9}{3}=378\).
Step 3
Exam Tip
विशेष वस्तुओं में से (0) या (1) चुनी जा सकती है। कुल \(\binom{3}{0}\binom{9}{4}+\binom{3}{1}\binom{9}{3}=378\) है।
Login to save your score, XP, coins and progress. Login
(8) खिलाड़ियों में से (3) खिलाड़ियों का चयन करना है। एक कप्तान पहले से तय है और उसे चुना नहीं जाना है। कितने तरीके हैं?
From (8) players (3) players are to be selected. One captain is already fixed and must not be selected. How many ways are there?
#combinations
#class11
#medium
A (35)
B (45)
C (56)
D (84)
Explanation opens after your attempt
Step 1
Concept
After removing the captain (7) players remain. Hence there are \(\binom{7}{3}=35\) ways.
Step 2
Why this answer is correct
The correct answer is A. (35). After removing the captain (7) players remain. Hence there are \(\binom{7}{3}=35\) ways.
Step 3
Exam Tip
कप्तान को हटाने पर (7) खिलाड़ी बचते हैं। इसलिए \(\binom{7}{3}=35\) तरीके हैं।
Login to save your score, XP, coins and progress. Login
(7) 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 (7) different toys?
#combinations
#class11
#medium
A (32)
B (63)
C (64)
D (127)
Explanation opens after your attempt
Step 1
Concept
The number of odd selections is \(2^{7-1}=64\). In exams even and odd selections are equal.
Step 2
Why this answer is correct
The correct answer is C. (64). The number of odd selections is \(2^{7-1}=64\). In exams even and odd selections are equal.
Step 3
Exam Tip
विषम चयन की संख्या \(2^{7-1}=64\) होती है। परीक्षा में सम और विषम चयन बराबर होते हैं।
Login to save your score, XP, coins and progress. Login
(6) 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 (6) different coins?
#combinations
#class11
#medium
A (16)
B (32)
C (31)
D (64)
Explanation opens after your attempt
Step 1
Concept
An even selection means (0), (2), (4), or (6) coins. The total is \(\binom{6}{0}+\binom{6}{2}+\binom{6}{4}+\binom{6}{6}=32\).
Step 2
Why this answer is correct
The correct answer is B. (32). An even selection means (0), (2), (4), or (6) coins. The total is \(\binom{6}{0}+\binom{6}{2}+\binom{6}{4}+\binom{6}{6}=32\).
Step 3
Exam Tip
सम संख्या का चयन (0), (2), (4), (6) सिक्कों का होगा। कुल \(\binom{6}{0}+\binom{6}{2}+\binom{6}{4}+\binom{6}{6}=32\) है।
Login to save your score, XP, coins and progress. Login
(10) अलग-अलग कार्डों में से (4) कार्ड चुनने हैं जिनमें एक निश्चित कार्ड हो या दूसरा निश्चित कार्ड हो लेकिन दोनों न हों। कितने तरीके हैं?
From (10) distinct cards (4) cards are to be selected containing exactly one of two fixed cards. How many ways are there?
#combinations
#class11
#medium
A (112)
B (120)
C (126)
D (140)
Explanation opens after your attempt
Step 1
Concept
Choose (1) of the two fixed cards and (3) cards from the remaining (8). The ways are \(\binom{2}{1}\binom{8}{3}=112\).
Step 2
Why this answer is correct
The correct answer is A. (112). Choose (1) of the two fixed cards and (3) cards from the remaining (8). The ways are \(\binom{2}{1}\binom{8}{3}=112\).
Step 3
Exam Tip
दो निश्चित कार्डों में से (1) चुनें और बाकी (3) कार्ड (8) में से चुनें। तरीके \(\binom{2}{1}\binom{8}{3}=112\) हैं।
Login to save your score, XP, coins and progress. Login
(5) वरिष्ठ और (7) कनिष्ठ कर्मचारियों में से (4) लोगों की समिति बनानी है जिसमें वरिष्ठों की संख्या कनिष्ठों से कम हो। कितने तरीके हैं?
From (5) senior and (7) junior employees a committee of (4) is to be formed with fewer seniors than juniors. How many ways are there?
#combinations
#class11
#medium
A (245)
B (280)
C (315)
D (350)
Explanation opens after your attempt
Step 1
Concept
The number of seniors can be (0) or (1). The total is \(\binom{5}{0}\binom{7}{4}+\binom{5}{1}\binom{7}{3}=210\).
Step 2
Why this answer is correct
The correct answer is D. (350). The number of seniors can be (0) or (1). The total is \(\binom{5}{0}\binom{7}{4}+\binom{5}{1}\binom{7}{3}=210\).
Step 3
Exam Tip
वरिष्ठों की संख्या (0) या (1) हो सकती है। कुल \(\binom{5}{0}\binom{7}{4}+\binom{5}{1}\binom{7}{3}=210\) है।
Login to save your score, XP, coins and progress. Login
(8) प्रश्नों में से (5) प्रश्न चुनने हैं और अंतिम (3) प्रश्नों में से कम से कम (2) प्रश्न चुनने हैं। कितने चयन होंगे?
From (8) questions (5) are to be selected and at least (2) of the last (3) questions must be selected. How many selections are there?
#combinations
#class11
#medium
A (25)
B (31)
C (36)
D (40)
Explanation opens after your attempt
Step 1
Concept
The cases are choosing (2) or (3) from the last (3). The total is \(\binom{3}{2}\binom{5}{3}+\binom{3}{3}\binom{5}{2}=40\).
Step 2
Why this answer is correct
The correct answer is B. (31). The cases are choosing (2) or (3) from the last (3). The total is \(\binom{3}{2}\binom{5}{3}+\binom{3}{3}\binom{5}{2}=40\).
Step 3
Exam Tip
मामले अंतिम (3) में से (2) या (3) चुनने के हैं। कुल \(\binom{3}{2}\binom{5}{3}+\binom{3}{3}\binom{5}{2}=40\) है।
Login to save your score, XP, coins and progress. Login
(9) विद्यार्थियों में से (4) की टीम बनानी है जिसमें (2) विशेष विद्यार्थी दोनों शामिल हों या दोनों न हों। कितने तरीके हैं?
From (9) students a team of (4) is to be formed in which (2) special students are either both included or both excluded. How many ways are there?
#combinations
#class11
#medium
A (42)
B (56)
C (70)
D (84)
Explanation opens after your attempt
Step 1
Concept
If both are included there are \(\binom{7}{2}=21\) ways and if both are excluded there are \(\binom{7}{4}=35\) ways. The total is (56).
Step 2
Why this answer is correct
The correct answer is C. (70). If both are included there are \(\binom{7}{2}=21\) ways and if both are excluded there are \(\binom{7}{4}=35\) ways. The total is (56).
Step 3
Exam Tip
दोनों शामिल हों तो \(\binom{7}{2}=21\) और दोनों बाहर हों तो \(\binom{7}{4}=35\) तरीके हैं। कुल (21+35=56) है।
Login to save your score, XP, coins and progress. Login
(7) कुर्सियों में से (4) कुर्सियां चुननी हैं और (2) खराब कुर्सियों में से कोई न चुनी जाए। कितने तरीके हैं?
From (7) chairs (4) chairs are to be selected and none of (2) broken chairs is selected. How many ways are there?
#combinations
#class11
#medium
A (5)
B (10)
C (15)
D (20)
Explanation opens after your attempt
Step 1
Concept
After removing (2) broken chairs (5) good chairs remain. The ways to choose (4) are \(\binom{5}{4}=5\).
Step 2
Why this answer is correct
The correct answer is A. (5). After removing (2) broken chairs (5) good chairs remain. The ways to choose (4) are \(\binom{5}{4}=5\).
Step 3
Exam Tip
(2) खराब कुर्सियां हटाने पर (5) अच्छी कुर्सियां बचती हैं। (4) चुनने के तरीके \(\binom{5}{4}=5\) हैं।
Login to save your score, XP, coins and progress. Login
(12) बिंदुओं से त्रिभुज बनाने हैं। यदि (5) बिंदु एक सीध में हैं तो केवल उन्हीं (5) बिंदुओं से बनने वाले असफल चयन कितने हैं?
Triangles are to be formed from (12) points. If (5) points are collinear then how many failed selections come only from those (5) points?
#combinations
#class11
#medium
A \(\binom{5}{2}\)
B \(\binom{5}{3}\)
C \(\binom{12}{3}\)
D \(\binom{7}{3}\)
Explanation opens after your attempt
Correct Answer
B. \(\binom{5}{3}\)
Step 1
Concept
For a failed triangle (3) points are chosen from the (5) collinear points. So the number is \(\binom{5}{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\binom{5}{3}\). For a failed triangle (3) points are chosen from the (5) collinear points. So the number is \(\binom{5}{3}\).
Step 3
Exam Tip
असफल त्रिभुज के लिए (5) समरेखीय बिंदुओं में से (3) चुने जाते हैं। इसलिए संख्या \(\binom{5}{3}\) है।
Login to save your score, XP, coins and progress. Login
(11) संख्याओं में से (4) संख्याएं चुननी हैं जिनमें (2) निश्चित संख्याएं शामिल हों और (1) निश्चित संख्या शामिल न हो। कितने तरीके हैं?
From (11) numbers (4) numbers are to be selected with (2) fixed numbers included and (1) fixed number excluded. How many ways are there?
#combinations
#class11
#medium
A (21)
B (28)
C (36)
D (45)
Explanation opens after your attempt
Step 1
Concept
The (2) numbers are fixed and (1) is excluded. The remaining (2) numbers are chosen from (8) in \(\binom{8}{2}=28\) ways.
Step 2
Why this answer is correct
The correct answer is B. (28). The (2) numbers are fixed and (1) is excluded. The remaining (2) numbers are chosen from (8) in \(\binom{8}{2}=28\) ways.
Step 3
Exam Tip
(2) संख्याएं तय हैं और (1) हट गई है। बाकी (2) संख्याएं (8) में से \(\binom{8}{2}=28\) तरीकों से चुनी जाएंगी।
Login to save your score, XP, coins and progress. Login
(6) विषयों में से (3) विषय चुनने हैं लेकिन गणित और भौतिकी दोनों साथ में नहीं चुने जा सकते। कितने तरीके हैं?
From (6) subjects (3) subjects are to be selected but mathematics and physics cannot both be selected together. How many ways are there?
#combinations
#class11
#medium
A (12)
B (14)
C (16)
D (18)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{6}{3}=20\) and if both special subjects are included the third subject is chosen in (4) ways. Hence (20-4=16).
Step 2
Why this answer is correct
The correct answer is C. (16). Total ways are \(\binom{6}{3}=20\) and if both special subjects are included the third subject is chosen in (4) ways. Hence (20-4=16).
Step 3
Exam Tip
कुल \(\binom{6}{3}=20\) हैं और दोनों विशेष विषय साथ हों तो तीसरा विषय (4) तरीकों से चुनेगा। इसलिए (20-4=16) है।
Login to save your score, XP, coins and progress. Login
(7) अलग-अलग अक्षरों में से (3) अक्षर चुनने हैं जिनमें (a) हो लेकिन (b) न हो। कितने तरीके हैं?
From (7) distinct letters (3) letters are to be selected containing (a) but not (b). How many ways are there?
#combinations
#class11
#medium
A (6)
B (10)
C (15)
D (20)
Explanation opens after your attempt
Step 1
Concept
The letter (a) is fixed and (b) is excluded so choose the remaining (2) letters from (5). The ways are \(\binom{5}{2}=10\).
Step 2
Why this answer is correct
The correct answer is B. (10). The letter (a) is fixed and (b) is excluded so choose the remaining (2) letters from (5). The ways are \(\binom{5}{2}=10\).
Step 3
Exam Tip
(a) तय है और (b) हट गया है इसलिए बाकी (2) अक्षर (5) में से चुनेंगे। तरीके \(\binom{5}{2}=10\) हैं।
Login to save your score, XP, coins and progress. Login
(4) लाल, (5) नीली और (6) हरी गेंदों में से प्रत्येक रंग की (1) गेंद चुनने के कितने तरीके हैं?
From (4) red, (5) blue, and (6) green balls, how many ways are there to choose (1) ball of each color?
#combinations
#class11
#medium
A (60)
B (90)
C (120)
D (180)
Explanation opens after your attempt
Step 1
Concept
One ball is chosen from each color. The ways are \(\binom{4}{1}\binom{5}{1}\binom{6}{1}=120\).
Step 2
Why this answer is correct
The correct answer is C. (120). One ball is chosen from each color. The ways are \(\binom{4}{1}\binom{5}{1}\binom{6}{1}=120\).
Step 3
Exam Tip
हर रंग से (1) गेंद चुनी जाएगी। तरीके \(\binom{4}{1}\binom{5}{1}\binom{6}{1}=120\) हैं।
Login to save your score, XP, coins and progress. Login
(9) फलों में से (4) फल चुनने हैं लेकिन (3) विशेष फलों में से ठीक (1) चुना जाए। कितने तरीके हैं?
From (9) fruits (4) fruits are to be selected but exactly (1) of (3) special fruits is chosen. How many ways are there?
#combinations
#class11
#medium
A (45)
B (60)
C (75)
D (90)
Explanation opens after your attempt
Step 1
Concept
Choose (1) from the special fruits and (3) from the remaining (6). The ways are \(\binom{3}{1}\binom{6}{3}=60\).
Step 2
Why this answer is correct
The correct answer is B. (60). Choose (1) from the special fruits and (3) from the remaining (6). The ways are \(\binom{3}{1}\binom{6}{3}=60\).
Step 3
Exam Tip
विशेष फलों में से (1) और बाकी (6) में से (3) चुनेंगे। तरीके \(\binom{3}{1}\binom{6}{3}=60\) हैं।
Login to save your score, XP, coins and progress. Login
(8) मित्रों में से (4) को यात्रा के लिए चुनना है और दो मित्रों में से कम से कम एक जाना चाहिए। कितने तरीके हैं?
From (8) friends (4) are to be selected for a trip and at least one of two friends must go. How many ways are there?
#combinations
#class11
#medium
A (45)
B (55)
C (60)
D (70)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{8}{4}=70\) and if both special friends do not go then \(\binom{6}{4}=15\). Hence (70-15=55).
Step 2
Why this answer is correct
The correct answer is B. (55). Total ways are \(\binom{8}{4}=70\) and if both special friends do not go then \(\binom{6}{4}=15\). Hence (70-15=55).
Step 3
Exam Tip
कुल \(\binom{8}{4}=70\) हैं और दोनों विशेष न जाएं तो \(\binom{6}{4}=15\) हैं। इसलिए (70-15=55) है।
Login to save your score, XP, coins and progress. Login
(10) उम्मीदवारों में से (3) पुरस्कार विजेताओं का चयन करना है और पुरस्कार समान हैं। कितने तरीके हैं?
From (10) candidates (3) prize winners are to be selected and the prizes are identical. How many ways are there?
#combinations
#class11
#medium
A (120)
B (360)
C (720)
D (1000)
Explanation opens after your attempt
Step 1
Concept
The prizes are identical so only selection is needed. The number of ways is \(\binom{10}{3}=120\).
Step 2
Why this answer is correct
The correct answer is A. (120). The prizes are identical so only selection is needed. The number of ways is \(\binom{10}{3}=120\).
Step 3
Exam Tip
पुरस्कार समान हैं इसलिए केवल चयन होगा। तरीकों की संख्या \(\binom{10}{3}=120\) है।
Login to save your score, XP, coins and progress. Login
(5) सफेद और (6) काली गेंदों में से (4) गेंदें चुननी हैं जिनमें रंग समान न हों। कितने तरीके हैं?
From (5) white and (6) black balls (4) balls are to be selected such that the colors are not all the same. How many ways are there?
#combinations
#class11
#medium
A (300)
B (310)
C (315)
D (320)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{11}{4}=330\). Removing all-white \(\binom{5}{4}=5\) and all-black \(\binom{6}{4}=15\) gives (310).
Step 2
Why this answer is correct
The correct answer is D. (320). Total ways are \(\binom{11}{4}=330\). Removing all-white \(\binom{5}{4}=5\) and all-black \(\binom{6}{4}=15\) gives (310).
Step 3
Exam Tip
कुल \(\binom{11}{4}=330\) हैं। सभी सफेद \(\binom{5}{4}=5\) और सभी काली \(\binom{6}{4}=15\) हटाने पर (310) मिलते हैं।
Login to save your score, XP, coins and progress. Login
(6) शिक्षकों और (8) छात्रों में से (5) लोगों का समूह बनाना है जिसमें कम से कम (2) शिक्षक हों। कितने तरीके हैं?
From (6) teachers and (8) students a group of (5) people is to be formed with at least (2) teachers. How many ways are there?
#combinations
#class11
#medium
A (1456)
B (1554)
C (1620)
D (1728)
Explanation opens after your attempt
Step 1
Concept
Total ways are \(\binom{14}{5}=2002\). Remove (0) teacher \(\binom{8}{5}=56\) and (1) teacher \(\binom{6}{1}\binom{8}{4}=420\). The answer is (1526).
Step 2
Why this answer is correct
The correct answer is B. (1554). Total ways are \(\binom{14}{5}=2002\). Remove (0) teacher \(\binom{8}{5}=56\) and (1) teacher \(\binom{6}{1}\binom{8}{4}=420\). The answer is (1526).
Step 3
Exam Tip
कुल \(\binom{14}{5}=2002\) हैं। (0) शिक्षक \(\binom{8}{5}=56\) और (1) शिक्षक \(\binom{6}{1}\binom{8}{4}=420\) हटाएं। उत्तर (1526) है।
Login to save your score, XP, coins and progress. Login
(13) विद्यार्थियों में से (5) चुनने हैं ताकि (3) विशेष विद्यार्थियों में से कोई भी शामिल न हो। कितने तरीके हैं?
From (13) students (5) are to be selected so that none of (3) special students is included. How many ways are there?
#combinations
#class11
#medium
A (252)
B (300)
C (792)
D (1287)
Explanation opens after your attempt
Step 1
Concept
After removing (3) special students (10) students remain. Hence there are \(\binom{10}{5}=252\) ways.
Step 2
Why this answer is correct
The correct answer is A. (252). After removing (3) special students (10) students remain. Hence there are \(\binom{10}{5}=252\) ways.
Step 3
Exam Tip
(3) विशेष विद्यार्थियों को हटाने पर (10) विद्यार्थी बचते हैं। इसलिए \(\binom{10}{5}=252\) तरीके हैं।
Login to save your score, XP, coins and progress. Login
(1) से (15) तक की संख्याओं में से (2) विषम और (1) सम संख्या चुनने के कितने तरीके हैं?
From numbers (1) to (15), how many ways are there to choose (2) odd and (1) even number?
#combinations
#class11
#medium
A (196)
B (210)
C (224)
D (280)
Explanation opens after your attempt
Step 1
Concept
There are (8) odd numbers and (7) even numbers. The ways are \(\binom{8}{2}\binom{7}{1}=196\).
Step 2
Why this answer is correct
The correct answer is A. (196). There are (8) odd numbers and (7) even numbers. The ways are \(\binom{8}{2}\binom{7}{1}=196\).
Step 3
Exam Tip
विषम संख्याएं (8) और सम संख्याएं (7) हैं। तरीके \(\binom{8}{2}\binom{7}{1}=196\) हैं।
Login to save your score, XP, coins and progress. Login
(1) से (12) तक की संख्याओं में से (3) संख्याएं चुननी हैं जिनमें सभी सम हों। कितने तरीके हैं?
From numbers (1) to (12), (3) numbers are to be selected and all must be even. How many ways are there?
#combinations
#class11
#medium
A (10)
B (15)
C (20)
D (24)
Explanation opens after your attempt
Step 1
Concept
There are (6) even numbers from (1) to (12). Hence there are \(\binom{6}{3}=20\) ways.
Step 2
Why this answer is correct
The correct answer is C. (20). There are (6) even numbers from (1) to (12). Hence there are \(\binom{6}{3}=20\) ways.
Step 3
Exam Tip
(1) से (12) तक (6) सम संख्याएं हैं। इसलिए \(\binom{6}{3}=20\) तरीके हैं।
Login to save your score, XP, coins and progress. Login
\(A=\{1,2,3,4,5,6,7,8\}\) के कितने (4)-तत्व उपसमुच्चय (2) और (3) दोनों को शामिल नहीं करते?
How many (4)-element subsets of \(A=\{1,2,3,4,5,6,7,8\}\) do not contain both (2) and (3) together?
#combinations
#class11
#medium
A (50)
B (55)
C (60)
D (70)
Explanation opens after your attempt
Step 1
Concept
Total subsets are \(\binom{8}{4}=70\) and those containing both (2), (3) are \(\binom{6}{2}=15\). Hence (70-15=55).
Step 2
Why this answer is correct
The correct answer is B. (55). Total subsets are \(\binom{8}{4}=70\) and those containing both (2), (3) are \(\binom{6}{2}=15\). Hence (70-15=55).
Step 3
Exam Tip
कुल \(\binom{8}{4}=70\) हैं और (2), (3) दोनों हों तो \(\binom{6}{2}=15\) हैं। इसलिए (70-15=55) है।
Login to save your score, XP, coins and progress. Login
कितने (3)-तत्व उपसमुच्चय \(A=\{1,2,3,4,5,6,7\}\) से बनाए जा सकते हैं जिनमें (1) शामिल हो?
How many (3)-element subsets can be formed from \(A=\{1,2,3,4,5,6,7\}\) that contain (1)?
#combinations
#class11
#medium
A (10)
B (12)
C (15)
D (20)
Explanation opens after your attempt
Step 1
Concept
The element (1) is already chosen so the remaining (2) elements are chosen from (6). The number of ways is \(\binom{6}{2}=15\).
Step 2
Why this answer is correct
The correct answer is C. (15). The element (1) is already chosen so the remaining (2) elements are chosen from (6). The number of ways is \(\binom{6}{2}=15\).
Step 3
Exam Tip
(1) पहले से चुना है इसलिए बाकी (2) तत्व (6) में से चुने जाएंगे। तरीकों की संख्या \(\binom{6}{2}=15\) है।
Login to save your score, XP, coins and progress. Login
\(\binom{10}{4}-\binom{9}{4}\) का मान क्या है?
What is the value of \(\binom{10}{4}-\binom{9}{4}\)?
#combinations
#class11
#medium
A (84)
B (126)
C (210)
D (336)
Explanation opens after your attempt
Step 1
Concept
\(\binom{10}{4}=210\) and \(\binom{9}{4}=126\). The difference is (84).
Step 2
Why this answer is correct
The correct answer is B. (126). \(\binom{10}{4}=210\) and \(\binom{9}{4}=126\). The difference is (84).
Step 3
Exam Tip
\(\binom{10}{4}=210\) और \(\binom{9}{4}=126\) हैं। अंतर (84) है।
Login to save your score, XP, coins and progress. Login
\(\binom{8}{3}+\binom{8}{4}\) पास्कल पहचान से किसके बराबर है?
Using Pascal's identity \(\binom{8}{3}+\binom{8}{4}\) is equal to which expression?
#combinations
#class11
#medium
A \(\binom{9}{4}\)
B \(\binom{9}{3}\)
C \(\binom{8}{7}\)
D \(\binom{16}{7}\)
Explanation opens after your attempt
Correct Answer
A. \(\binom{9}{4}\)
Step 1
Concept
By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{9}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\binom{9}{4}\). By Pascal's identity \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\). Hence the answer is \(\binom{9}{4}\).
Step 3
Exam Tip
पास्कल पहचान से \(\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}\) होता है। इसलिए उत्तर \(\binom{9}{4}\) है।
Login to save your score, XP, coins and progress. Login
यदि \(\binom{n}{1}+\binom{n}{2}=36\) है तो (n) का मान क्या है?
If \(\binom{n}{1}+\binom{n}{2}=36\), what is the value of (n)?
#combinations
#class11
#medium
A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
It gives (n+\frac{n(n-1)}{2}=36). Putting (n=8) gives (8+28=36).
Step 2
Why this answer is correct
The correct answer is B. (8). It gives (n+\frac{n(n-1)}{2}=36). Putting (n=8) gives (8+28=36).
Step 3
Exam Tip
यह (n+\frac{n(n-1)}{2}=36) देता है। (n=8) रखने पर (8+28=36) मिलता है।
Login to save your score, XP, coins and progress. Login
