Combinations Straight Flush in Texas Hold'em Poker

Question

On the Wikipedia page about Poker probabilities for 7-card poker hands (Link) it says the following for the Royal Flush: \begin{equation*} \binom{4}{1} \binom{47}{2} \end{equation*} First we choose 1 of the 4 suits, then the next 5 cards are given and then we could have anything for the last two cards.

For the Straight Flush it says this: \begin{equation*} \binom{9}{1} \binom{4}{1} \binom{46}{2} \end{equation*} First we choose the top rank of our straight (5, 6, ..., King), then we choose the suit and finally the last two cards could be anything. BUT why do we have 2 out of 46 and not 47 as for the Royal Flush?

Answer 1

Your link says

The Ace-high straight flush or royal flush is slightly more frequent (4324) than the lower straight flushes (4140 each) because the remaining two cards can have any value; a King-high straight flush, for example, cannot have the Ace of its suit in the hand (as that would make it ace-high instead).

So the two unused cards can be anything except the five making up the straight flush and the card of the flush suit immediately above the straight flush which would make it a different hand. So there are $52-5-1=46$ other cards and you choose two of these.

There is nothing above an Ace so in a royal flush you choose two from $52-5=47$.