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Sum

In an examination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.

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#### Solution

Given that question number 1 and 2 are compulsory

∴ The remaining questions are 5 – 2 = 3

Total number of questions to be attempted = 4 questions 1 and 2 are compulsory

So only 2 questions are to be done out of 3 questions

Therefore number of ways = ^{3}C_{2}

= ^{3}C_{3–2}

= 3 ......`[∴ ""^nC_r = ""^nC_(n - r)]`

Hence, the required number of ways = 3.

Concept: Combination

Is there an error in this question or solution?

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