How many 9-card hands contain four cards of the same value?

Question

There are 52 cards, 4 suits, each suits have 13 differences value.

step1 -> There are 13 difference case, 13C1 = 13

step2 -> Without 4 cards, the remaining cards is 48.And we need to pick up other 5 cards. 48C5

Then 13 * 48C5 is the answer for How many 9-card hands contain four cards of the same value. But the answer is incorrect, so i need some hints,Thanks!.

Answer 1

In $13\cdot \binom{48}{5} $ you counted two times the cases where there are TWO distinct values with four cards, so you have to subtract $\binom{13}{2}\cdot (52-8)$.