Two-Suit Five-Card Hand Count
A standard deck has 52 cards divided into 4 suits of 13 cards each. How many 5-card hands contain exactly 2 cards from one suit and exactly 3 cards from a different suit?
Note: the two suits play different roles (one contributes 2 cards, the other contributes 3), so pay attention to whether the order of suit selection matters.
Hints
- Think carefully: does swapping which suit contributes 2 cards versus 3 cards produce a different hand?
- Since the two suits play different roles, use an ordered selection (permutation) rather than a combination when picking the two suits.
- Use $P(4,2) = 12$ for the ordered suit choice, then multiply by $\binom{13}{3}$ and $\binom{13}{2}$ for the card selections within each suit.
Worked Solution
How to Think About It: The key question is whether swapping the two suits gives a different hand. Yes -- a hand with 3 hearts and 2 spades is a different hand from 3 spades and 2 hearts. So we need an ordered selection of 2 suits from 4, not just an unordered pair. Once the suits are assigned, the card selection within each suit is independent and unordered.
Quick Estimate: There are 4 suits, so there are $4 \times 3 = 12$ ordered ways to pick the "3-card suit" and the "2-card suit". Within each suit, we are choosing from 13 cards. The cards-per-suit counts are $\binom{13}{3} = 286$ and $\binom{13}{2} = 78$. Rough product: $12 \times 286 \times 78 \approx 12 \times 22{,}000 \approx 264{,}000$. The exact answer should be close to that.
Approach: Use the multiplication principle: choose the ordered pair of suits, then choose cards from each suit independently.
Formal Solution:
Step 1: Choose the suits. We need to assign one suit the role of "3-card suit" and another the role of "2-card suit". Since these roles are distinct, we use a permutation: $$P(4,2) = 4 \times 3 = 12$$
Step 2: Choose cards from the 3-card suit. Select 3 cards from 13 in that suit: $$\binom{13}{3} = 286$$
Step 3: Choose cards from the 2-card suit. Select 2 cards from 13 in the other suit: $$\binom{13}{2} = 78$$
Step 4: Multiply. $$P(4,2) \cdot \binom{13}{3} \cdot \binom{13}{2} = 12 \times 286 \times 78 = 267{,}696$$
Answer: $\boxed{267{,}696}$
Intuition
The trap here is treating suit selection as unordered -- picking clubs and hearts is not the same as picking hearts and clubs when one is the 3-card suit and the other is the 2-card suit. Whenever the objects you are selecting play asymmetric roles, you need a permutation, not a combination. This is a common source of double-counting errors in counting problems.
In practice, this kind of reasoning -- carefully tracking which objects are distinguishable and which are not -- comes up constantly in combinatorics. A good sanity check: if you used $\binom{4}{2} = 6$ instead of $P(4,2) = 12$, you would get $6 \times 286 \times 78 = 133{,}848$, exactly half the correct answer. That factor of 2 is the ordered pair you missed.