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Calculating Combinations


Here you will learn how to calculate basic combinations , which are arrangements of items that ignore the order of the items.


Suppose there are fifteen students in a classroom, and you are looking to build a team of five students for a tug-of-war.  How many different groups of five students could you create?

After the following lesson, you should have no problem solving combination question of this type.

Watch This

http://youtu.be/bCxMhncR7PU Khan Academy – Combinations


In other lessons, we have spent quite a bit of time learning how to count permutations , which are arrangements of items that take order into account.  Now we will be considering combinations , which are groups of items only concerned with which items are in the arrangement, regardless of the order.

Naturally there are fewer unique combinations of a particular group of items than there are permutations, since there may be many permutations of the same single combination of items.  In fact, the formula for counting combinations without repetition is actually just a modification of the formula for permutations where the denominator is increased.  Take a look at the two formulas side-by-side:

\begin{align*}Permutations: \ \frac{n!}{(n-r)!} \qquad Combinations: \ \frac{n!}{(n-r)! \times r!} \end{align*}

Where  \begin{align*}n =\end{align*} number of objects to choose from, and  \begin{align*}r =\end{align*} the number chosen

Notice how the combination formula is the same, with only the addition of the  \begin{align*}r!\end{align*} in the denominator of the fraction.  This makes sense conceptually, given that we need to reduce the number of arrangements that are the same items in a different order.

In summary, to calculate the number of combinations possible from  \begin{align*}n\end{align*} items chosen \begin{align*}r\end{align*} at a time, use the combination formula: \begin{align*}\frac{n!}{(n-r)! \times r!}\end{align*}

Example A

The local ice cream shop carries twelve flavors.  How many different three-scoop bowls are possible?


There are twelve flavors, so \begin{align*}n=12\end{align*} .  We are creating bowls with three scoops each, so \begin{align*}r=3\end{align*} :

\begin{align*}& \qquad \qquad \qquad \qquad \ \ \frac{12!}{(12-3)! \times 3!}\\ & \qquad \qquad \qquad \qquad \qquad \frac{12!}{9! \times 3!}\\ & \frac{12 \times 11 \times 10 \times {\color{red}\bcancel{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}}{{\color{red}\bcancel{(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}}(3 \times 2 \times 1)} \ (\text{cancel terms})\\ & \qquad \qquad \qquad \qquad \ \ \qquad \frac{1320}{6}\\ & \qquad \qquad \qquad \qquad 220 \ different \ flavors\end{align*}

Example B

If there are thirteen students in a room, how many ways can a group of five be formed from them?


There are thirteen students to choose from, so  \begin{align*}n = 13\end{align*}

We are choosing five at a time, so  \begin{align*}r = 5\end{align*}

\begin{align*}& \qquad \quad \frac{13!}{(13-5)! \times 5!}\\ & \qquad \qquad \frac{13!}{(8!)(5!)}\\ & \quad \ \frac{13 \times 12 \times 11 \times 10 \times 9}{5!}\\ & \qquad \qquad \frac{154440}{120}\\ & 1287 \ \text{possible groups of five}\end{align*}

Example C

If your student council has seven members, how many ways are there to reach a majority vote?


There are seven members to choose from, so \begin{align*}n = 7\end{align*} .  A majority vote is either four, five, six, or seven votes, so we need to calculate the number of combinations for each \begin{align*} r = 4, 5, 6\end{align*} , and 7, and add them together.

a. Four votes:

\begin{align*}& \qquad \frac{7!}{(7-4)!\times4!}\\ & \frac{7 \times 6 \times 5 \times 4 {\color{red}\bcancel{ \times 3 \times 2 \times 1}}}{{\color{red}\bcancel{(3 \times 2 \times 1)}}(4 \times 3 \times 2 \times 1)}\\ & \qquad \quad \frac{840}{24}=35\end{align*}

b. Five votes:

\begin{align*}& \quad \frac{7!}{(7-5)! \times (5!)}\\ & \frac{7 \times 6 \times 5 \times 4 \times 3}{5 \times 4 \times 3 \times 2 \times 1}\\ & \quad \ \frac{2520}{120}=21\end{align*}

c. Six votes:

\begin{align*}& \quad \ \frac{7!}{(7-6)! \times 6!}\\ & \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\\ & \qquad \frac{5040}{720}=7\end{align*}

d. Seven votes:

\begin{align*}& \frac{7!}{(7-7)! \times 7!}\\ & \quad \ \ \frac{7!}{7!}=1\end{align*}

\begin{align*}\text{Total} = 35+21+7+1 = 64\end{align*}  ways to reach a majority vote

Concept Problem Revisited

Suppose there are fifteen students in a classroom, and you are looking to build a team of five students for a tug-of-war. How many different groups of five students could you create?

This is a combinations problem requiring us to choose items five at a time from a pool of 15, which can be notated as 15C5 (read as "fifteen, choose five").  Use the combinations formula with n=15 and r=5:

\begin{align*}\text{Combinations} & = \frac{n!}{(n-r)! \times r!} \\ & = \frac{ 15! }{ \big( (15-5)! \times 5! \big) } \\ & = \frac{ 15! }{ 10! \times 5! } \\ & = \frac{ 15 \times 14 \times 13 \times 12 \times 11 }{ 5! } \\ & = \frac{ 360,360 }{ 120 }\end{align*}

3,003 different teams


A combination is a unique collection of items.

A permutation is a unique arrangement of items.

 Guided Practice

  1. How many ways can six cars be chosen from a lot containing twenty-one cars?
  2. How many unique groups of four can me made from nine pieces of candy?
  3. How many unique five-card hands are possible using a standard deck of cards?


1. There are twenty-one cars to choose from,  \begin{align*}n = 21\end{align*}

There are six cars in each group, \begin{align*}r = 6\end{align*}

\begin{align*}& \qquad \frac{21!}{((21-6)! \times 6!)}\\ & \frac{21 \times 20 \times 19 \times 18 \times 17 \times 16}{6!}\\ & \quad 54264 \ combinations\end{align*}

2. Nine pieces to choose from, \begin{align*}n = 9\end{align*} , and four in each group, \begin{align*}r = 4\end{align*} :

\begin{align*}& \frac{9!}{(9-4)! \times 4!}\\ & \frac{9 \times 8 \times 7 \times 6}{4!}\\ & 756 \ groups\end{align*}

3. There are fifty-two card in a standard deck, \begin{align*}n = 52\end{align*} , and five in each hand, \begin{align*}r = 5\end{align*} :

\begin{align*}& \quad \quad \ \frac{52!}{(52-5)! \times 5!}\\ & \quad \frac{52 \times 51 \times 50 \times 49 \times 48}{5!}\\ & 62,375,040 \ possible \ hands\end{align*}


  1. In the ball closet, there are 6 basketballs, footballs, and baseballs. You go in and pick 2 items at random. Assuming all the balls are marked differently, how many different groups of balls can you end up with?
  2. You go to a store and buy 4 various items from 6 distinct choices. How many different possibilities were there?
  3. Find the number of combinations of choosing 5 items from 9 distinct items.
  4. Find the number of combinations of choosing 6 items from 9 distinct items.
  5. You have a pile of 7 distinctly different coins. How many ways can you take 2 coins from the pile?
  6. You are dealt a hand of 4 cards from a deck of 6 cards. Find the number of possible hands you can get
  7. You hire 2 out of 9 people to build a garden. Find the number of possible ways of choosing your workers.
  8. Find the number of combinations of choosing 4 items from 7 distinct items.
  9. Find the number of combinations of choosing 5 items from 5 distinct items.
  10. How many five-card hands are possible from a deck of twenty-six cards?
  11. How many unique 7-person combinations are possible using 32 people?
  12. If Pandi has nineteen chickens in her coop, how many unique groups of six chickens are possible from them?
  13. If one of her chickens lays colored eggs, every egg unique, how many four-color groups could she make from a dozen eggs?




Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.


A permutation is an arrangement of objects where order is important.

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