Question

If $15$ books are arranged in a row then the number of ways of choosing $4$ books such that all the four books are not together is

• A. $1351$
• B. $1352$
• C. $1353$
• D. $1365$

Answer 1

The correct option is C $1353$

Total no. of ways of selecting $4$ books such that all the $4$ books are not together = total no. of ways - total ways when all the books are together

Total no. of ways = ${}^{15}{C}_4$
Total no. of ways such that all $n$ books are consecutive $=\left(\right(a_1,a_2,a_3,a_4),(a_2,a_3,a_4,a_5),........,(a_{n-3},a_{n-2},a_{n-1},a_n)$
$=n-3$

For $n=15$, the number of ways $=12$

​​​​​Hence the required number of ways$={}^{15}{C}_4-12=1353$