Combinations

If you need to pick 3 players from a team of 9, how many different groups could you choose? Questions like this are what combinations are for.

A combination is a selection of items from a larger set where the order doesn't matter. Think of choosing pizza toppings — pepperoni, mushrooms, and olives is the same order as olives, mushrooms, and pepperoni. You end up with the same pizza either way, so sequence is irrelevant. The same logic applies to picking people for a team, choosing books to read, or selecting any group where the arrangement doesn't change the outcome.

That's the key distinction: combinations count what you picked, not how you arranged it. Picking John, Fred, and Bill is the same combination as picking Fred, Bill, and John.

The Formula

To count combinations, use this formula:

$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

The notation \(\binom{n}{r}\) means "choose \(r\) items from a group of \(n\)." You'll also see this written as \(nCr\) or \(C(n,r)\) — all the same thing.

The \(!\) symbol means factorial: the product of that number and every positive integer below it. So \(4! = 4 \times 3 \times 2 \times 1 = 24\), and \(6! = 720\).

Worked Examples

Example 1: How many ways can you choose 3 players from a team of 9?

Here \(n = 9\) and \(r = 3\):

$$\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3! \cdot 6!}$$

Notice that most of this cancels. You don't need to multiply out all of \(9!\) — the \(6!\) in the denominator wipes out everything in \(9!\) except the top three factors:

$$= \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$$

There are 84 different ways to pick 3 players from a team of 9.

Example 2: A traveler wants to visit 6 of the 50 U.S. states (in no particular order). How many different trips are possible?

$$\binom{50}{6} = \frac{50!}{6! \cdot 44!} = \frac{50 \times 49 \times 48 \times 47 \times 46 \times 45}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$$

$$= \frac{11{,}441{,}304{,}000}{720} = 15{,}890{,}700$$

Over 15 million possible trips. That's why combinations get large fast.

Combinations vs. Permutations

Combinations and permutations use similar formulas, but the key difference is order. In a combination, the sequence doesn't matter — a group of three people is the same group regardless of who's listed first. In a permutation, order matters — first place, second place, and third place are three different outcomes.

When you're deciding which one to use, ask yourself: would rearranging the selected items give you a different result? If yes, it's a permutation. If no, it's a combination.

Give These a Shot

  1. How many ways can you choose 2 students from a class of 10?

  2. A pizza place lets you pick any 4 toppings from a list of 12. How many different pizzas can you order?

  3. A committee of 5 is being formed from a group of 8 people. How many possible committees are there?

Answers:

  1. \(\binom{10}{2} = \dfrac{10 \times 9}{2 \times 1} = 45\)

  2. \(\binom{12}{4} = \dfrac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \dfrac{11{,}880}{24} = 495\)

  3. \(\binom{8}{5} = \dfrac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1} = \dfrac{6{,}720}{120} = 56\)

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