# The Formula for Combinations

And now, finally, let’s get to the formula for combinations. The math in my last post got a little tricky toward the end, with the strange exclamation mark notation floating around. So let’s recap permutations without that notation.

4
× 3
× 3 × 2
× 3 × 2 × 1
÷ (1 × 2 × 3)
÷ (1 × 2)
÷ 1

You should see that, to traverse a tree diagram, we multiply by the tree branches, cumulatively, to move right, and then divide by those branches—again, cumulatively—to move left. The formula for the number of permutations of 2 cards chose from 4, $$\mathtt{\frac{4!}{(4-2)!}}$$, tells us to multiply all the way to the right, to get 24 in the numerator and then divide two steps to the left (divide by $$\mathtt{(4-2)!}$$ or 2) to get 12 permutations of 2 cards chosen from 4.

Combinations

An important point about the above is that the number of permutations of $$\mathtt{r}$$ cards chosen from $$\mathtt{n}$$ cards, $$\mathtt{_{n}P_r}$$, is a subset of the number of permutations of $$\mathtt{n}$$ cards, $$\mathtt{n!}$$ The tree diagram shows $$\mathtt{n!}$$ and contained within it are $$\mathtt{_{n}P_r}$$.

Combinations of $$\mathtt{r}$$ items chosen from $$\mathtt{n}$$, denoted as $$\mathtt{_{n}C_r}$$, are a further subset. That is, $$\mathtt{_{n}C_r}$$ are a subset of $$\mathtt{_{n}P_r}$$. In our example of 2 cards chosen from 4, $$\mathtt{_{n}P_r}$$ represents the first two columns of the tree diagram combined. In those columns, we have, for example, the permutations JQ and QJ. But these two permutations represent just one combination. The same goes for the other pairs in those columns. Thus, we can see that to get the number of combinations of 2 cards chosen from 4, we take $$\mathtt{_{n}P_r}$$ and divide by 2. So, $\mathtt{\frac{4!}{(4-2)!}\div 2=\frac{4!}{(4-2)!\cdot2}}$

What about combinations of 3 cards chosen from 4? That’s the first 3 columns combined. Now the repeats are, for example, JQK, JKQ, QJK, QKJ, KJQ, KQJ. Which is 6. Noticing the pattern? For $$\mathtt{_{4}C_2}$$, we divide $$\mathtt{_{4}P_2}$$ further by 2! For $$\mathtt{_{4}C_3}$$, we divide $$\mathtt{_{4}P_3}$$ further by 3! We’re dividing (further) by $$\mathtt{r!}$$

When you think about it, this makes sense. We need to collapse every permutation of $$\mathtt{r}$$ cards down to 1 combination. So we divide by $$\mathtt{r!}$$ Here, finally then, is the formula for combinations: $\mathtt{_{n}C_r=\frac{n!}{(n-r)!r!}}$