Question

In how many different ways a student can choose 8 question out of 11 question ,if 3 question are complusory ?​

Answer 1

Answer:

studied permutation and combination during preparation. It had been a long time since I solved a question on that. But might be I correct here, I think this is an easy question.

Simply, either you choose 8 questions out of 10 to answer or 2 questions out of 10 not to answer. The answer is same and it is 10C2. And this is solved as 10!/(8!×2!). The answer is 45. Always when you get a question to select r things for n things, the answer is nPr. Which is again n!/{r!×(n-r)!}.