# Ways to make a straight in poker

A deck is 4 suits of 13 cards each. The ace is both low and high so that is why you have 10 ways to make a straight.

A flush is all cards of the same suite. There are two hands (flush and full house) between straight and straight-flush so this is why they are broken out.

I am not able to follow the calculation for number of way to make a straight (excluding straight flush) from here

$\binom{10}{1} \binom{4}{1}^5 - \binom{10}{1} \binom{4}{1}$

The part I don't get is the $\binom{4}{1}^5$
Where does that come from?

Is that 4 suits and 5 cards? Even if so I don't understand.

I am not a keen poker player but I believe this refers to the possibility of selecting a card from each of the four suits. For each of the five cards in the straight, this can be done in $\binom{4}{1} = 4$ ways . For instance, for a straight of 1,2,3,4,5 there are four fives among which you can choose. And likewise for the rest of the cards.
Also, a word about the $\binom{10}{1}$ factor. This accounts for the number of ways one can select five adjacent cards out of the thirteen, there are nine ways to do so, plus the possibility of selecting the last four cards and having a high ace, which still counts as a straight I hear. In all, 10 ways.
• +1 And "$-{10 \choose 1}{4 \choose 1}$" discounts the straight flush, and its subset, the royal flush. – Antoni Parellada Jan 30 '16 at 20:37