Question

A man has $3$ friends. If $N$ is the number of ways he can invite one friend everyday for dinner on $6$ successive nights so that no friend is invited more than $3$ times, then the value of $N/170$ is

• A. 3.00
• B. 3.0
• C. 3

Answer 1

Answer: (A), (B), (C)

$\left(i\right)$ He can invite $2$ friends three times each
Lets select first those $2$ friends in ${}^3{C}_2$ ways
Now these two friends each three time can be invited on $6$ days in $\frac{6!}{3!3!}$
Thus total number of ways 2 friends can be invited three times $={}^3{C}_2\times \frac{6!}{3!3!}$

$\left(ii\right)$ Another possibility is that he invites all three friends $2$ times each
Then number of ways $=\frac{6!}{2!2!2!}$

$\left(iii\right)$ One more possibility is that he invites one friend only once, one two times and one three times.
Then the friend who has to be invited thrice can be selected as${}^3{C}_1$, now the friend who has to be invited twice can be selected as${}^2{C}_1$
total number of ways $={}^3{C}_1\times {}^2{C}_1\times \frac{6!}{2!3!}$

Hence total number of ways
$={}^3{C}_2\times \frac{6!}{3!3!}+\frac{6!}{2!2!2!}+{}^3{C}_1\times {}^2{C}_1\times \frac{6!}{2!3!}=510$
$\therefore N/170=3$