Question

How many ways are there to invite one of three friends for dinner on $6$ successive nights such that to friend is invited more than three times ?

• A. $\frac{6\times 6!}{1!2!3!}+3\times \frac{6!}{3!3!}+\frac{6!}{2!2!2!}$
• B. $\frac{6\times 6!}{1!2!3!}+6\times \frac{6!}{3!3!}+\frac{6!}{2!2!2!}$
• C. $\frac{6\times 6!}{1!2!3!}+\frac{6!}{3!3!}+\frac{6!}{2!2!2!}$
• D. $\frac{3\times 6!}{1!2!3!}+3\times \frac{6!}{3!3!}+\frac{6!}{2!2!2!}$

Answer 1

The correct option is A $\frac{6\times 6!}{1!2!3!}+3\times \frac{6!}{3!3!}+\frac{6!}{2!2!2!}$

Case I : Invite $2$ friends for $3$ night each

No.of ways $={}^3{C}_2{}^6{C}_3{}^3{C}_3\to$

$\downarrow$

choosing $3$ digits out of $6$

choosing $3$ nights out of remaining $3$ nights

Case II : Invite $1$ friend $3$ times, another friend $2$ times & third friend $1$ time , No of ways $=3!{}^6{C}_3{}^3{C}_2{}^1{C}_1$

Case III: Invite all three friends $2$ times

No. of ways $={}^6{C}_2{}^4{C}_2{}^2{C}_2$

Total no of ways $={}^3{C}_2{}^6{C}_3{}^3{C}_3+3!{}^6{C}_2{}^1{C}_1+{}^6{C}_2{}^4{C}_2{}^2{C}_2$

$=\frac{3\times 6!}{3!6!}+\frac{6\times 6!}{3!3!}\times 3+\frac{6!}{2!4!}\frac{4!}{2!2!}$

$=\frac{6\times 6!}{3!2!}+3\times \frac{6!}{3!3!}+\frac{6!}{2!2!2!}$

Option $A$ is correct