Question

A man has $7$ letters for his $7$ friends. The letter are kept in the envelopes at random. The number of ways in which exactly $3$ letters are going to correct envelope and rest $4$ letters are going to the wrong envelopes is

• A. $210$
• B. $315$
• C. $350$
• D. $420$

Answer 1

The correct option is B $315$

The number of ways of $3$ letters going to correct envelopes is ${}^7{C}_3=35$

The number of ways in which $4$ letters are going to wrong envelopes is $4!(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!})=9$

Hence, the required number of ways is $35\times 9=315$