Poker hand - difference from any full house vs a specific full house

Question

I'm having trouble understanding the solution for the first part of the question below:

Problem: In a poker hand consisting of 5 cards, find the probability P(A) of holding 2 aces and 3 jacks.

My Solution: I started solving this using the same logic as the solution for any full house, but like this:

Part 1

• for the 2 aces: 13C1 * 4C2 (because out of the 13 ranks, I am interested on only 1 - the Ace)
• for the 3 jacks: 13C1 * 4C3 (same reason as above, I am interested only in the appearance of the Jacks, that's why I use 13).
• This would give me 13*6*13*4 = 4,056

Part 2

• All probabilities for a 5 hand poker in a 52 card: 52C5 = 2,598,960

Then: P(A) = 4,056/2,598,960 = 0.00156, which is wrong.

Solution from book: I split in 2 parts too, where part 2 is similar to what I did.

Part 1:

• The number of ways of being dealt 2 aces from 4 cards is 4C2 = 6. And the number of ways of being dealt 3 jacks from 4 cards is 4C3 = 4. By the multiplication rule, there are n = (6)(4) = 24 hands with 2 aces and 3 jacks.

Part 2

• 52C5 = 2,598,960.

Therefore, the probability of getting 2 aces and 3 jacks in a 5-card poker hand is P(A) = 24/2,598,960 = 0.9 x 10^-5.

Question: Why is the first part calculated like that, considering only 4 cards and not the whole rank of 13 cards like in calculating any full house?

Thanks.

Answer 1

When you include the $\binom{13}{1}$ term, you are saying that you can make ANY one choice out of thirteen possibilities.

What you actually want is to say that there is only one choice that works -- hence you want to multiply by $1$, not by $13$.