SOLUTION: A student is required to work exactly 6 problems from a 10 problem exam and must work exactly 3 of the first 4 problems. In how many ways can the six problems be chosen?
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Question 360156: A student is required to work exactly 6 problems from a 10 problem exam and must work exactly 3 of the first 4 problems. In how many ways can the six problems be chosen?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
If he has to work exactly 6 out of 10 and 3 out of the first 4, then he must work exactly 3 out of the last 6. Hence, the answer to this question is to be found in the sum of the number of ways to select 3 things from 4 things plus the number of ways to select 3 things from 6 things:
!}\ +\ \frac{6!}{3!(6\,-\,3)!})
You can do your own arithmetic.
John
My calculator said it, I believe it, that settles it
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