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Question

A boy has $3$ library tickets and $8$ books of his interest in the library of these $8,$ he does not want to borrow maths part $2$ unless maths part $1$ is borrowed? In how many ways can he choose the three books to be borrowed?

(a) $41$

(b) $51$

(c) $52$

(d) $61$

(a) $41$

(b) $51$

(c) $52$

(d) $61$

2016-12-21 12:17:28

Keyur thakkar
2 Answers

__Case 1: 3 books without including maths part 1 and part 2__

$6\text{C}3=20$ ways

__Case 2: (maths part 1) and (two books)__

$7\text{C}2=21$ ways

Required number of ways

$=20+21=41$

Number of ways in which three books can be selected without any conditions

$=8\text{C}3=56$.

Now find out number of ways in which three books can be selected in which maths part $2$ must appear and part $1$ must not appear. For this, remove maths part $1$ and part $2$ first and select two books from the remaining $6$ books. This can be done in $6\text{C}2=15$ ways.

Required number of ways

$=56-15=41$

$=8\text{C}3=56$.

Now find out number of ways in which three books can be selected in which maths part $2$ must appear and part $1$ must not appear. For this, remove maths part $1$ and part $2$ first and select two books from the remaining $6$ books. This can be done in $6\text{C}2=15$ ways.

Required number of ways

$=56-15=41$