Right and Wrong Permutations and Combinations Methods And Pointers That Will Make You Ace These On Your CSET Exam

Notice that in this combinations problem, order does not matter. The number we plug in for n is 6 because there are a total of 6 letters. The number we plug in for k is 1 because we are taking only 1 letter at a time. We can now plug these values into our combinations formula:

Notice that both the numerator and denominator have the number of factors indicated by the lower index, which in this case is 1. The numerator has 1 factor beginning with the upper index 6 and going backwards (although in this problem, because the lower index is 1, we only go 1 place). The denominator has 1 factor beginning with 1 and going forward (again, we only go out 1 number).

We can visualize this answer by doing what the problem tells us, writing out all the combinations and seeing if they do equal 6:
1 = a
2 = b
3 = c
4 = d
5 = e
6 = f

Yes, all the combinations of abcdef, taken 1 letter at a time, does equal our original answer of 6.

What if we were to do the same problem, but instead of taking 1 letter at a time, we take 2 letters at a time?

Let’s visualize our answer by actually writing out all the combinations of abcdef taken 2 letters at a time and seeing if they do equal 15:
1 = ab
2 = ac
3 = ad
4 = ae
5 = af
6 = bc
7 = bd
8 = be
9 = bf
10 = cd
11 = ce
12 = cf
13 = de
14 = df
15 = ef

Yes, all the combinations of abcdef, taken 2 letters at a time, does equal our original answer of 15.