Non negative integral solution of x, y, z, w.
Case 1 : x = 0. then
y + z + w = 14 . This has one to one correspondence with selection with repetition or distributing r similar ball to n distinguishable boxes which is given by $\binom{n + r - 1}{r}$ .
So $\binom{14 + 3 - 1}{14}$ = $\binom{14 + 3 - 1}{2}$ = 16! / (2! * 14!) = 8 * 15 = 120.
Case 2 : x = 1, then
y + z + w = 14 - 3 = 11. So no of ways doing this is $\binom{13}{2}$ = 13! / (2! * 11!) = 13 * 6 = 78.
Case 3 : x = 2 then
y + z + w = 8 . No. of ways doing this is $\binom{10}{2}$ = 5 * 9 = 45.
Case 4 : x = 3 then
y + z + w = 5. So , $\binom{7}{2}$ = 21 ways
Case 5: x = 4 then
y + z + w = 2. So , $\binom{4}{2}$ = 6.ways
Case 6. x = 5 . It can't be possible.
So Answer : 120 + 78 + 45 + 21 + 6 = 270.
