Deck Of Cards - 12 Card Hands

Question

Problem: Find the number of twelve-card hands containing four cards in each of three suits.

My solution:

1. pick a suit : 4 ways.
2. choose 4 cards from the 13 in that suit: ${^{13}}{C}_{4}$ ways.
3. pick another suit: 3 ways.
4. choose 4 cards from the 13 in the new suit: ${^{13}}{C}_{4}$ ways.
5. pick the final suit: 2 ways.
6. choose 4 cards from the 13 in the final suit: ${^{13}}{C}_{4}$ ways.

Hence, total number of ways is $4\cdot{^{13}}{C}_{4}\cdot3\cdot{^{13}}{C}_{4}\cdot2\cdot{^{13}}{C}_{4} = {^{4}}{P}_{3}\cdot\left({^{13}}{C}_{4}\right)^3 $ total hands.

This is the incorrect answer. Why?

Answer 1

You get the incorrect answer because you overcount. If you pick the suits of spades, clubs and then hearts, your method counts that as different form picking first hearts, then spades, then clubs, even though they result in the same hands.

Actually, you count each hand exactly six times, once for each order of the three chosen suits, so you should get the correct answer by just dividing by $6$. If you want the correct answer from the start, you could begin by picking three suits, which may be done in $^4C_3$ ways. Then you pick four cards from the first suit, four cards from the second suit, and four cards from the third suit. In other words, your $P$ should've been a $C$.