the binomial coefficient (n choose k)
/ by-NOH-mee-ul /
The number written C(n,k), or 'n choose k', answers one question with two faces. Combinatorially it is the number of ways to choose k items from n when order does not matter — exactly the combination count. Algebraically the very same number is the COEFFICIENT that sits in front of x^k when you expand (1 + x)^n. It is remarkable that 'how many ways to pick a subset' and 'an algebra coefficient' are the same thing, and that coincidence is the doorway to a lot of mathematics.
The formula is C(n,k) = n! / (k! (n-k)!), defined for whole numbers 0 <= k <= n (and taken as 0 if k is outside that range). Two properties make it friendly. First, symmetry: C(n,k) = C(n, n-k), because choosing the k items to take is the same as choosing the n-k items to leave behind — C(10,3) = C(10,7). Second, the boundary values C(n,0) = C(n,n) = 1 (one way to take nothing, one way to take everything). Why does it appear in (1+x)^n? Expanding means choosing, from each of the n factors, either the 1 or the x; you get x^k exactly when you picked the x from k of the factors, and there are C(n,k) ways to choose which k.
These numbers are everywhere: they fill Pascal's triangle, they ARE the coefficients in the binomial theorem, and in probability they give the chance of exactly k successes in n independent trials, P = C(n,k) p^k (1-p)^(n-k). The sum of a whole row, C(n,0) + C(n,1) + ... + C(n,n), equals 2^n — which just says the total number of subsets of an n-element set is 2^n, since each element is either in or out.
C(5,2) = 5!/(2!3!) = 10, and by symmetry C(5,3) is also 10. Check the subset-total: row 5 is 1, 5, 10, 10, 5, 1, which sums to 32 = 2^5 — the number of all subsets of a 5-element set.
C(n,k) counts subsets AND is the coefficient of x^k in (1+x)^n; C(n,k)=C(n,n-k).
C(n,k) is always a whole number even though the formula divides factorials — the k! and (n-k)! always divide n! exactly. Treat it as one indivisible quantity, not as three separate factorials to compute apart.