@freemo

Might benefit from a discussion of why nCk = n!/(k!(n-k)!). I don’t use combinatorics often enough to have it memorised, so I re-derive the nCk and nPk formulae every time I need them. Here’s my reasoning:

• If you want to arrange n objects, you can choose any of n for the first, any of the n-1 remaining for the second, … down to one for the nth. So n objects can be arranged in n! = n(n-1)...1 ways.
• If you want to choose k of n objects, you can do this by ordering the n objects and taking the first k. So initially nCk = n!
• However, you don’t care about ordering of the chosen objects. If you choose two of {1, 2, 3, 4}, the orderings {1, 2, 3, 4} and {2, 1, 3, 4} should not be treated as distinct. Since there are k! ways to order the k objects you chose, you need to divide that out. Now we have nCk = n!/k!
• Equally, you don’t care about ordering of the non-chosen objects. In the above example, {1, 2, 3, 4} and {1, 2, 4, 3} should not be treated as distinct. There are n-k of these objects, so (n-k)! ways to order them, which again should be divided out. Finally, we arrive at nCk = n!/(k!(n-k)!)