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

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 (2, 2, 2) or (3, 3, 0)
Hence, the total number of ways

$=\frac{6!}{1!2!3!}\times 3!+\frac{6!}{2!2!2!}\times \frac{3!}{3!}+\frac{6!}{3!3!0!}\times \frac{3!}{2!}$

$=360+90+60=510$