Question

The number of ways in which $5$ balls numbered $1-5$ can be placed in $5$ boxes numbered as $1-5$ (each box can accomodate one ball only) when

• A. all the balls should be kept on correspnding numbered boxes is $5$
• B. one ball is not kept in correspnding numbered box is $1$
• C. three balls are not kept in correspnding numbered boxes is $20$
• D. four balls are not kept in correspnding numbered boxes is $45$

Answer 1

Answer: (C), (D)

The correct options are
C three balls are not kept in correspnding numbered boxes is $20$
D four balls are not kept in correspnding numbered boxes is $45$

$\left(i\right)$The number of ways when all the balls should be kept on corresponding numbered boxes will be $1$
since there is only one way to keep all the balls in correspnding numbered boxes

$\left(ii\right)$The number of ways when one ball is not kept in corresponding numbered box will be $0$
(Atleast two balls will be wrong boxes)

$\left(iii\right)$When three balls are not kept in corresponding numbered boxes.
The number of ways of selecting $2$ balls going to corresponding numbered boxes ${}^5{C}_2=10$
The number of ways in which $3$ balls not going to corresponding numbered boxes is $3!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!})=2$
Hence, the required answer is $10\times 2=20$

$\left(iv\right)$When four balls are not kept in corresponding numbered boxes.
The number of ways of selecting $1$ ball going to corresponding numbered box is ${}^5{C}_1=5$
The number of ways in which $4$ balls not going to corresponding numbered boxes is $4!(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})=9$
Hence, the required answer is $5\times 9=45$