Asked by  | 1st Dec, 2009, 12:36: PM

Expert Answer:

If he borrows chemistry I then for the remaining two books out of 7 = 7C2 = 7!/(2!5!) = 21 ways

and if he don't borrow chemistry I then, he'll not borrow chemistry II,

The combinations are 6C3 = 6!/(3!3!) = 20

Total 41 different ways he can borrow three books.





Answered by  | 1st Dec, 2009, 11:02: PM

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