# Harris, Combinatorics - Five-card hands - Solution verification

I would like someone to go over my solution and verify if my reasoning is correct. This is problem 8 from Harris, Combinatorics.

Compute the number of ways to deal each of the following five-card hands in poker.

(a) Straight: the values of the cards form a sequence of consecutive integers. A jack has a value $11$, a queen $12$ and a king $13$. An ace may have a value of $1$ or $14$, so $A2345$ and $10JQKA$ are both straights, but $KA234$ is not. Furthermore, the cards in a straight cannot all be of the same suit(a flush).

A straight will be one of the following $10$ sequences :

$(A,2,3,4,5)\\(2,3,4,5,6)\\(3,4,5,6,7)\\(4,5,6,7,8)\\(5,6,7,8,9)\\(6,7,8,9,10)\\(7,8,9,10,J)\\(8,9,10,J,Q)\\(9,10,J,Q,K)\\(10,J,Q,K,A)$

Let $E$ be the event that a straight (with cards from any suit) are obtained.

Let $S$ be the event that a straight will all cards in the same suit are obtained.

$n(E)=10\cdot{{4\choose1}}^{5}$

$n(S)=10\cdot{{4\choose1}}$

The required number of hands = $n(E)-n(S)=10,200$

Aside :

Why doesn't this add up step by step?

(a) 2 cards of suit A, 3-cards of suit B: $10\cdot{4\choose1}\cdot {3\choose1}$

(b) 2 cards of suit A, 2-cards of a suit B, 1 card of suit C: $10\cdot{4\choose1}\cdot {3\choose1}\cdot {2\choose1}$

(c) 2 cards of suit A, 1-card of a suit B, 1 card of suit C, 1 card of suit D: $10\cdot{4\choose1}\cdot {3\choose1}\cdot {2\choose1}\cdot {1\choose1}$

• Yep, your working's correct. – Parcly Taxel Oct 30 '16 at 5:08
• @ParclyTaxel, why doesn't the step by step approach add up? – Quasar Oct 30 '16 at 5:32
• You fail to take into account how the suits can be distributed over the ranks. – Parcly Taxel Oct 30 '16 at 5:34
• @ParclyTaxel, it would be real helpful, if you could elicit, how that works... :) – Quasar Oct 30 '16 at 5:40

Your list isn't complete, and you haven't computed properly.

Counting it as [ Choose suits ] $\times$ [ Place them ] for one straight, cases are:

2 suits, AAAAB or AAABB: $\binom41\binom31(\frac{5!}{4!1!} + \frac{5!}{3!2!}) = 180$

3 suits, AAABC or AABBC: $\binom41\binom32(\frac{5!}{3!1!1!} + \frac{5!}{2!2!1!}) = 600$

4 suits, AABCD: $\binom41\binom33\frac{5!}{2!1!1!1!1!} = 240$

Add and multiply by $10$ for the $10$ straights !

I have avoided multinomials, and written permutations in full, if you are conversant with multinomials, you can abridege, just write each case as the product of $2$ multinomials

• Thanks, this inclusion approach is just mind-blowing for me. :) One - why are arranging them? Coz, no matter what order you are dealt the cards in, the hand is the same right? Isnt AAAAB is equivalent to AAABA? Two - In 3 suits, for AABBC - shouldn't that be ${4 \choose 1}{3 \choose 1}{2 \choose 1}$? – Quasar Oct 30 '16 at 6:44
• 1. If the straight is (3,4,5,6,7), AAAAB is not equivalent to AAABA because B gets alloted to a different rank. 2. If you do that, you would be considering $\color{red}{\heartsuit\heartsuit\diamondsuit\diamondsuit}\spadesuit$ and $\color{red}{\diamondsuit\diamondsuit\heartsuit\heartsuit}\spadesuit$ as different choices of suit. – true blue anil Oct 30 '16 at 7:13
• @trueblueanil - nice use of $\color{red}{\heartsuit\diamondsuit}\spadesuit\clubsuit$ :) – hypergeometric Oct 30 '16 at 14:47
• @hypergeometric: Thanks ! :) – true blue anil Oct 30 '16 at 17:17