The Binomial Theorem says that, for a positive integer #n#,

#(x+b)^n= ""_nC_0 x^n + ""_nC_1 x^(n-1)b+""_nC_2 x^(n-2)b^2 + ... + ""_nC_n b^n#

This can be succinctly written as the sum

#(x+b)^n= sum_(k=0)^n ""_nC_k x^(n-k)b^k#

To see why this formula works, let's use it on the binomial for this question, #(x+2)^4#. If we were to write out all the factors side-by-side, we'd get

#(x+2)^4=(x+2)(x+2)(x+2)(x+2)#

Multiplying this out means making all possible products of one term from each binomial, and adding these products together. Sort of like FOIL-ing to the next level. (In FOIL-ing, there are #color(blue)2# binomials, so there will be #2^color(blue)2=4# terms; with #color(green)4# binomials, there will be #2^color(green)4=16# terms.)

After (tediously) multiplying it all out, we'll get some terms with #x^4#, some with #x^3 * 2#, some with #x^2 * 2^2#, and so on. But how many of each?

Since there is only #""_4C_0=1# way to pair up all four #x#'s, we'll only get a single #x^4# term.

There are #""_4C_1=4# ways to make an #x^3 * 2# term—using the #2# from binomials 1, 2, 3, and 4 (multiplied with the #x#'s from the others).

Continuing this way, there will always be #""_4C_k# ways to make an #x^(n-k) * 2^k# term. So this is where the combination notation in the Binomial Theorem comes from. For #k=0,...,4#, the #x^(n-k) * 2^k# bit (from the sum above) represents the different terms we can make, and the #""_4C_k# bit is the number of ways to make that term.

Thus, the binomial expansion of #(x+2)^4# is

#""_4C_0 x^4 + ""_4C_1 x^3 * 2 + ""_4C_2 x^2 * 2^2 + ""_4C_3 x * 2^3 + ""_4C_4 * 2^4 #

#=1x^4+4x^3(2)+6x^2(4)+4x(8)+1(16)#

#=x^4+8x^3+24x^2+32x+16#.