Class 11 Mathematics Permutations and Combinations Combinations Medium Level 63

(11) विद्यार्थियों में से (5) की टीम बनानी है जिसमें (2) विशेष विद्यार्थी दोनों शामिल हों या दोनों बाहर हों। कितने तरीके हैं?

From (11) 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?

Explanation opens after your attempt
Correct Answer

C. (280)

Step 1

Concept

If both are included there are \(\binom{9}{3}=84\) ways and if both are excluded there are \(\binom{9}{5}=126\) ways. The total is (210).

Step 2

Why this answer is correct

The correct answer is C. (280). If both are included there are \(\binom{9}{3}=84\) ways and if both are excluded there are \(\binom{9}{5}=126\) ways. The total is (210).

Step 3

Exam Tip

दोनों शामिल हों तो \(\binom{9}{3}=84\) और दोनों बाहर हों तो \(\binom{9}{5}=126\) तरीके हैं। कुल (84+126=210) है।

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(11) विद्यार्थियों में से (5) की टीम बनानी है जिसमें (2) विशेष विद्यार्थी दोनों शामिल हों या दोनों बाहर हों। कितने तरीके हैं? / From (11) 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?

Correct Answer: C. (280). Explanation: दोनों शामिल हों तो \(\binom{9}{3}=84\) और दोनों बाहर हों तो \(\binom{9}{5}=126\) तरीके हैं। कुल (84+126=210) है। / If both are included there are \(\binom{9}{3}=84\) ways and if both are excluded there are \(\binom{9}{5}=126\) ways. The total is (210).

Which concept should I revise for this Mathematics MCQ?

If both are included there are \(\binom{9}{3}=84\) ways and if both are excluded there are \(\binom{9}{5}=126\) ways. The total is (210).

What exam hint can help solve this Mathematics question?

दोनों शामिल हों तो \(\binom{9}{3}=84\) और दोनों बाहर हों तो \(\binom{9}{5}=126\) तरीके हैं। कुल (84+126=210) है।