The following has been extracted from the Combinations article in the Hold ‘em Mathology blog on Tumblr. It may be useful for those comfortable with high school math,
https://holdemmathology.tumblr.com/p...r-calculations
Combinations are the number of ways hands or boards can be dealt, without regard to order. Qc 8d for example can occur with the Q being the first or second card. But even though that is two possibilities, disregarding the order makes for just one combination.
The combinatorial formula for dealing r cards out of a total of n cards, or choosing r things out of a total of n things, is
C(n,r) = n!/(r! * (n-r)!), 0 =< r <= n, n > 0
where x! = x * (x-1)*(x-2)* … *3 * 2 * 1 ( e.g., 4! = 4 * 3 * 2 * 1 = 24)
Other notations for combos include nCr and n Choose r. Note that C(n,0) = 1 for any n.
Example: How many pairs are there for a particular rank. A rank has 4 cards. A pair is gotten from any two cards of that rank. So, n=4 and r=2. Therefore, the number of combos is 4!/(2!*2!) = (4*3*2*1)/(2*1 *2*1) = 6. Of course, for small n, combos can be gotten directly by counting. For a pair, the enumerated 6 combos are cd, ch, cs, dh, ds, hs.