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How many different selections of 5 books can be made from 12 different books if, Two particular books are always selected?

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#### Solution

Total number of books = 12

Number of books to be selected = 5

Given Two books are always selected.

Remaining number of books to be selected = 3

The number of ways of selecting the remaining 3 books from the remaining 10 books = 10C_{3 }

= `(10!)/(3! xx (10 - 3)!)`

= `(10!)/(3! xx 7!)`

= `(10 xx 9 xx 8 xx 7!)/(3! xx 7!)`

= `(10 xx 9 xx 8)/(3!)`

= `(10 xx 9 xx 8)/(3 xx 2 xx 1)`

= 5 × 3 × 8

= 120 ways

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