A lottery system consists of 40 balls numbered 1 through 40. In the Big 5, five balls are selected from these 40 balls. You need to match all five numbers to win the cash prize. How many ways can 5 balls from a selection of 40 numbers be drawn?
Combinations
Combinations of a subset of a larger set of objects refer to the number of ways we can choose items in any order. For comparison, look at the table below to see when order matters and when order doesn’t matter.
Combinations  Permutations 







The simplest way to describe the difference between a combination and a permutation is to say that in a combination the order doesn’t matter. The members of a committee could be selected in any order but the officers in a club are assigned a specific position and therefore the order does matter. Be careful of the use of these words in the real world as they are sometimes misused. For example, a locker combination. The ways to select and order the three numbers for a locker combination is not actually a combination, but a permutation since the order does matter.
Solve the following problem using combinations
How many ways can we choose three different flavors of ice cream from a selection of 15 flavors to place in a bowl?
First, does order matter in the bowl? When we created an ice cream cone with three scoops in an earlier concept the order did matter but here it does not. Let’s work from the example of the ice cream cone. We determined the number of permutations of a subset of three flavors from the total 15 flavors using the formula: \begin{align*}\frac{n!}{(nr)!}=\frac{15!}{(153)!}=\frac{15!}{12!}=\frac{15\times14\times13\times{\color{red}\cancel{12!}}}{{\color{red}\cancel{12!}}}=2730\end{align*}. Now that order doesn’t matter, this number includes the \begin{align*}3!\end{align*} ways to arrange each combination of 3 flavors. We can divide 2,730 by \begin{align*}3!\end{align*} to determine the number of combinations: \begin{align*}\frac{2730}{3\times2\times1}=\frac{2730}{6}=455\end{align*}.
The notation and formula for combinations can be written as: \begin{align*}\dbinom{n}{r}={_nC}_r=C^n_r=\frac{n!}{r!(nr)!}\end{align*}, where \begin{align*}n\end{align*} represents the number of elements in the set and \begin{align*}r\end{align*} represents the number of elements in the subset.
Evaluate the following expressions:
1) \begin{align*}\dbinom{8}{5}\end{align*}
\begin{align*}\dbinom{8}{5}=\frac{8!}{5!(85)!}=\frac{8\times7\times{\color{red}6}\times{\color{red}\cancel{5!}}}{{\color{red}\cancel{5!}}\times{\color{red}3}\times{\color{red}2}\times1}=\frac{8\times7}{1}=56\end{align*}.
2) \begin{align*}_8C_0\end{align*}
Type in 8 on the TI83 Graphing calculator, then MATH\begin{align*}\rightarrow\end{align*}PRB, select 3: \begin{align*}_nC_r\end{align*}. Now type in 0 and your screen should read 8 \begin{align*}_nC_r\end{align*} 0 before your press ENTER to get the answer 1.
3) \begin{align*} _8C_8\end{align*}
Type in 8 on the TI83 Graphing calculator, then MATH\begin{align*}\rightarrow\end{align*}PRB, select 3: \begin{align*}_nC_r\end{align*}. Now type in 8 and your screen should read 8 \begin{align*}_nC_r\end{align*} 8 before your press ENTER to get the answer 1.
4) \begin{align*}C^{10}_7\end{align*}
\begin{align*}C^{10}_7=\frac{10!}{7!(107)!}=\frac{10\times9\times8\times{\color{red}\cancel{7!}}}{{\color{red}\cancel{7!}}\times3!}=\frac{10\times{\color{red}\cancel{3}}\times3\times4\times{\color{red}\cancel{2}}}{{\color{red}\cancel{3}}\times{\color{red}\cancel{2}}}=120\end{align*}
5) Explain why the answers to 2 and 3 are the same.
In problem 2, we are looking at the ways to choose 0 items from 8 choices. There is only one way to do this. In problem 3 we are looking at the ways to choose 8 items from 8 choices. Well, the only way to do that is to choose all 8 items. So, there is only 1 way to choose zero items or all the items from a set.
All of the notations in problems 14 indicate that we should use the formula for a combination. We can use the graphing calculator to evaluate these as well. Problems 2 and 3 are set up in the form of the calculator notation so we will use the calculator to evaluate those two and the formula for the other two.
Solve the following problem using combination
How many ways can a team of five players be selected from a class of 20 students?
We can express this problem using the notation \begin{align*}\dbinom{20}{5}\end{align*} and then use the formula to evaluate.
\begin{align*}\dbinom{20}{5}=\frac{20!}{5!\times15!}=\frac{{\color{red}\cancel{20}}\times19\times{\color{red}\cancel{6}}\times3\times17\times16\times{\color{red}\cancel{15!}}}{{\color{red}\cancel{5\times4}}\times{\color{red}\cancel{3\times2}}\times{\color{red}\cancel{15!}}}=15,504.\end{align*}
Examples
Example 1
Earlier, you were asked how many ways can 5 balls from a selection of 40 numbers be drawn.
First, we need to determine whether order matters. The winning numbers 5, 10, 15, 20, and 25 are the same as the winning numbers 25, 5, 20, 10, 15, so order does not matter.
Therefore, we use the combinations formula \begin{align*}\dbinom{n}{r}=\frac{n!}{r!(nr)!}\end{align*}
\begin{align*}\frac{40!}{5!(405)!}\\ \frac{40!}{5!35!}\\ \frac {40\cdot39\cdot38\cdot37\cdot36\cdot35!}{5\cdot4\cdot3\cdot2\cdot1\cdot 35!}\\ \frac{78,960,960}{120} = 258,008\end{align*}
Therefore, there are 258,008 ways five winning balls can be drawn from the 40 numbers.
Evaluate the following using the formula for combinations of the calculator.
Example 2
\begin{align*}\dbinom{7}{5}\end{align*}
\begin{align*}7 \ _nC_r \ 5 = 21\end{align*}
Example 3
\begin{align*}_{20}C_{12}\end{align*}
\begin{align*}20 \ _nC_r \ 12 = 125,970\end{align*}
Example 4
\begin{align*}C^{15}_7\end{align*}
\begin{align*}15 \ _nC_r \ 7 = 6,435\end{align*}
Example 5
How many ways can a committee of three students be formed from a club of fifteen members?
\begin{align*}\dbinom{15}{3}=\frac{15!}{3!(153)!}=\frac{{\color{red}\cancel{3}}\times5\times{\color{red}\cancel{2}}\times7\times13\times{\color{red}\cancel{12!}}}{{\color{red}\cancel{3\times2}}\times{\color{red}\cancel{12!}}}=455\end{align*}.
Example 6
How many threetopping pizzas can be made if there are 10 topping choices?
\begin{align*}C^{10}_3=10 \ _nC_r \ 3 = 120\end{align*}.
Review
Evaluate the following combinations with or without a calculator.
 \begin{align*}_{13}C_{10}\end{align*}
 \begin{align*}C^{10}_6\end{align*}
 \begin{align*}\dbinom{18}{10}\end{align*}
 Explain why \begin{align*}_9C_5={_9C}_4=126\end{align*}.
 Decide whether the following situations are permutations and which are combinations.
 Ways to arrange students in a row.
 Ways to select a group of students.
 Ways to organize books on a shelf.
 Ways to select books to read from a larger collection.
 Ways to select three different yogurt flavors from a collection of ten flavors.
In each scenario described below, use either a combination or permutation as appropriate to answer the question.
 There are seven selections for appetizers on a caterer’s menu. How many ways can you select three of them?
 You only have time for seven songs on your workout playlist. If you have 10 favorites, how many ways can you select seven of them for the list? Now, how many ways can you select them in a particular order?
 How many ways can you select two teams of five players each from a group of ten players?
 At the local frozen yogurt shop a sundae comes with your choice of three toppings. If there are 12 choices for toppings, how many combinations of toppings are possible?
 How many ways can four people be selected from a group of 30 to serve on a committee? What if each of the four people was selected to fill a specific position on the committee?
 A soccer team has 20 players, but only 11 play at any one time.
 How many ways can the coach select a group of eleven players to start (disregard positions)?
 Now, of the eleven players on the field, one is a goalie, four play defense, three play midfield and three play offense. How many ways are there to assign the eleven players to these positions?
 Considering your answers to parts a and b, how many ways can the coach select eleven players and assign them positions on the field? Assume all players can play each position.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 12.5.