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7.9: Probability and Combinations

Difficulty Level: Basic Created by: CK-12
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Suppose you're at the local animal shelter, and you want to adopt two dogs. If there are 45 dogs from which to choose, how many different pairs of dogs could you adopt? What formula do you think you could use to calculate this number? If the dogs were chosen at random, would it be possible for you to find the probability of adopting a particular pair? 

Probability and Combinations

When the order of objects is not important and/or the objects are replaced, combinations are formed.

A combination is an arrangement of objects in no particular order.

Consider a sandwich with salami, ham, and turkey. The order in which we place the deli meat does not matter, as long as it’s on the sandwich.

There is only one way to stack the meat on the sandwich if the order does not matter. However, if the order mattered, there are 3 choices for the first meat, 2 for the second, and one for the last choice: \begin{align*}\underline{3} \cdot \underline{2} \cdot \underline{1}=6\end{align*}.

\begin{align*}Combination \neq Permutation\end{align*}

A combination of \begin{align*}n\end{align*} objects chosen \begin{align*}k\end{align*} at a time is expressed as \begin{align*}_nC_k\end{align*}.

\begin{align*}_nC_k =\frac{n!}{k!(n-k)!}= \binom{n}{k}\end{align*}

This is read “\begin{align*}n\end{align*} choose \begin{align*}k\end{align*}.”

Let's use combinations to solve the following problem:

How many ways can 8 students be chosen from a class of 21?

It does not matter how the eight students are chosen. Use the formula for combinations rather than permutations.

\begin{align*}_{21}C_8&=\frac{21!}{8!(21-8)!}\\ &=\frac{21!}{8!13!}\\ &=\frac{21\cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\cdot13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\\ &=\frac{21\cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot \cancel{13} \cdot \cancel{12} \cdot \cancel{11} \cdot \cancel{10} \cdot \cancel9 \cdot \cancel 8 \cdot \cancel 7 \cdot \cancel 6 \cdot \cancel 5 \cdot \cancel 4 \cdot \cancel 3 \cdot \cancel 2 \cdot \cancel 1}{ 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\cdot \cancel{13} \cdot \cancel{12} \cdot \cancel{11} \cdot \cancel{10} \cdot \cancel9 \cdot \cancel 8 \cdot \cancel 7 \cdot \cancel 6 \cdot \cancel 5 \cdot \cancel 4 \cdot \cancel 3 \cdot \cancel 2 \cdot \cancel 1}\\ &= 203,490\end{align*}

There are 203,490 different ways to choose eight students from 21.

Combinations on a Graphing Calculator

Just like permutations, most graphing calculators have the capability to calculate combinations. On the TI calculators, use these directions.

  • Enter the \begin{align*}n\end{align*}, or the total to choose from.
  • Choose the [MATH] button, directly below the [ALPHA] key. Move the cursor once to the left to see this screen:

  • Choose option #3, \begin{align*}_nC_r\end{align*}. Type in the \begin{align*}k\end{align*} value, the amount you want to choose.

Let's use a graphing calculator to solve the following problem:

Calculate \begin{align*}\binom{100}{4}\end{align*} using a graphing calculator.


Probability and Combinations

Combinations are used in probability similarly to permutations but combinations are used when there is a replacement of objects or the order does not matter.

Use combinations to solve the following problem:

Suppose you have ten marbles: four blue and six red. You choose three marbles without looking. What is the probability that all three marbles are blue?

\begin{align*}\text{Probability} \ (success) = \frac{number \ of \ ways \ to \ get \ success}{total \ number \ of \ possible \ outcomes}\end{align*}

There are \begin{align*}_4C_3\end{align*} ways to choose the blue marbles. There are \begin{align*}_{10}C_3\end{align*} total combinations.

\begin{align*}P(all \ 3 \ marbles \ are \ blue)= \frac{\binom{4}{3}}{\binom{10}{3}} = \frac{4}{120}=\frac{1}{30}\end{align*}

There is approximately a 3.33% chance that all three marbles drawn are blue.


Example 1

Earlier, you were told that there are 45 dogs at the shelter and you want to choose 2 of them to adopt. How many different pairs could you adopt? If the dogs were chosen at random, would it be possible for you to find the probability of adopting a particular pair?

Since we do not care what order the dogs are chosen in, this is a combination problem. We want to know how many groups of 2 we can choose from 45. 

\begin{align*}_{45}C_2 &= \frac{45!}{2!(45-2)!}\\ &=\frac{45!}{2!43!}\\ &=990\end{align*} 

There are 990 ways to adopt 2 of the 45 dogs. 

To find the probability that a particular pair of dogs is adopted, note that there is only 1 way to successfully pick that pair. There are 990 ways to pick any pair. Therefore, the probability is \begin{align*}\frac{1}{990}\end{align*}.

Example 2

The U.S. Senate is made up of 100 people, two per state. How many different four-person committees are possible?

This question does not care how the committee members are chosen; we will use the formula for combinations.

\begin{align*}\binom{100}{4}=\frac{100!}{4!(100-4)!}=3,921,225 \ ways\end{align*}

That is a lot of possibilities!

Example 3

Using the information from Example 2, what is the probability that the committee will only have members from two states?

If there are only members from two states, that means two are from one state and two are from another. This problem is simply about how many ways you can choose 2 states out of 50 states.

\begin{align*}\binom{50}{2}=\frac{50!}{2!(50-2)!}=1,225 \ ways\end{align*}

\begin{align*} P(\text{only two states represented})= \frac{1225}{3,921,225}=0.00031\end{align*}

The probability that only two states will be represented on the committee is 0.031%, which is a very small chance!


  1. What is a combination? How is it different from a permutation?
  2. How many ways can you choose \begin{align*}k\end{align*} objects from \begin{align*}n\end{align*} possibilities?
  3. Why is \begin{align*}_3C_9\end{align*} impossible to evaluate?

In 4 – 19, evaluate the combination.

  1. \begin{align*}\binom{12}{2}\end{align*}
  2. \begin{align*}\binom{8}{5}\end{align*}
  3. \begin{align*}\binom{5}{1}\end{align*}
  4. \begin{align*}\binom{3}{0}\end{align*}
  5. \begin{align*}\binom{9}{9}\end{align*}
  6. \begin{align*}\binom{9}{4}\end{align*}
  7. \begin{align*}\binom{20}{10}\end{align*}
  8. \begin{align*}\binom{19}{18}\end{align*}
  9. \begin{align*}\binom{20}{14}\end{align*}
  10. \begin{align*}\binom{13}{9}\end{align*}
  11. \begin{align*}_7C_3\end{align*}
  12. \begin{align*}_{11}C_5\end{align*}
  13. \begin{align*}_5C_4\end{align*}
  14. \begin{align*}_{13}C_9\end{align*}
  15. \begin{align*}_{20}C_5\end{align*}
  16. \begin{align*}_{15}C_{15}\end{align*}
  17. Your backpack contains 6 books. You select two at random. How many different pairs of books could you select?
  18. Seven people go out for dinner. In how many ways can 4 order steak, 2 order vegan, and 1 order seafood?
  19. A pizza parlor has 10 toppings to choose from. How many four-topping pizzas can be created?
  20. Gooies Ice Cream Parlor offers 28 different ice creams. How many two-scooped cones are possible, given that order does not matter?
  21. A college football team plays 14 games. In how many ways can the season end with 8 wins, 4 losses, and 2 ties?
  22. Using the marble situation from the Concept, determine the probability that the three marbles chosen are all red?
  23. Using the marble situation from the Concept, determine the probability that two marbles are red and the third is blue.
  24. Using the Senate situation from the Concept, how many two-person committees can be made using Senators?
  25. Your English exam has seven essays and you must answer four. How many combinations can be made?
  26. A sociology test has 15 true/false questions. In how many ways can you answer 11 correctly?
  27. Seven people are applying for two vacant school board positions; four are women, and three are men. In how many ways can these vacancies be filled ...
    1. With any two applicants?
    2. With only women?
    3. With one man and one woman?

Mixed Review

  1. How many ways can 15 paintings be lined along a wall?
  2. Your calculator gives an “Overload” error when trying to simplify \begin{align*}\frac{300!}{296!}\end{align*}. What can you do to help evaluate this fraction?
  3. Consider a standard six-sided die. What is the probability that the number rolled will be a multiple of 2?
  4. Solve the following system: The sum of two numbers is 70.6 and their product is 1,055.65. Find the two numbers.

Review (Answers)

To see the Review answers, open this PDF file and look for section 7.9. 

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combination Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
Permutation A permutation is an arrangement of objects where order is important.
Probability Probability is the chance that something will happen. It can be written as a fraction, decimal or percent.

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