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Question
How many different selections of 5 books can be made from 12 different books if, Two particular books are always selected?
Sum
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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 = 10C3
= `(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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Chapter 4: Combinatorics and Mathematical Induction - Exercise 4.3 [Page 187]
