Question

A man has three friends. The number of ways he can invite one friend everyday for dinner on six successive nights so that no friend is invited more than three times is

• A. $640$
• B. $320$
• C. $420$
• D. $510$

Answer 1

Answer: (D)

Let $x, y, z$ be the friends and $a, b, c$ denote the case when $x$ is invited $a$ times, $y$ is invited $b$ times, and $z$ is invited $c$ times.
Now, we have the following possibilities:
$(a, b, c) = (1,2,3)$ or $(3\, 3\, 0)$ or $(2 \,2 \,2)$
[grouping of $6$ days of week]
Hence, the total number of ways is
$\frac{6!}{1! 2! 3!} 3! + \frac{6!}{3! 3! 2!}3! + \frac{6!}{(2! 2! 2!)3!}3!$
$= 360 + 60 + 90 = 510$