Question

A man has three friends. The number of ways he can invite one friend everynight for dinner on six successive nights so that no friend is invited more than three times is $10N$, then $N=$

Answer 1

For the given case we have the following possibilities:
$(1,2,3)\text{or}(3,3,0)\text{or}(2,2,2)$
[grouping of $6$ days of week]
Hence, the total number of ways is
$\frac{6!}{1!2!3!}3!+\frac{6!}{3!3!2!}3!+\frac{6!}{(2!2!2!)3!}3!$
$=360+60+90=510$