Section 5.4: Combinations

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Learning Objectives

• Distinguish between a combination and a permutation.
• Find the number of combinations of n objects taken r at a time.
• Use the combination rule in conjunction with the fundamental counting principle.

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In the previous section on permutations, we developed systematic methods for counting arrangements where order matters. We learned to calculate the number of ways to arrange objects in sequences, select and arrange items from larger groups, and solve probability problems involving rankings, orderings, and positions. The key insight throughout our study of permutations was that changing the order of objects creates different, distinguishable outcomes, whether we were arranging books on a shelf, determining race finishing orders, or finding the number of ways a president and vice president can be selected from a group of people.

However, as we worked through various counting problems, you may have encountered situations where our permutation approach seemed to over-complicate the solution. Consider the difference between these two questions:

• "In how many ways can we select and arrange 3 students from a class of 20 to serve as president, vice-president, and secretary?" (This is a permutation problem because roles matter)
• "In how many ways can we select 3 students from a class of 20 to serve on a committee?" (This suggests a different type of counting problem)

In the second question, we don't care about the order in which we select the students or how we arrange them, we simply want to know how many different groups of 3 students we can form. If we used our permutation methods here, we would be counting each group of 3 students multiple times (once for each possible arrangement), leading to an incorrect and inflated count.

This brings us to an entirely different category of counting problems: situations where order does not matter. In these scenarios, we're interested in selecting or choosing objects from a group, but we don't care about the sequence or arrangement of our selections. Whether we choose students A, B, and C or students C, A, and B for our committee, we end up with the same committee. This type of selection problem requires a different mathematical approach than the arrangement problems we solved with permutations.

Combinations provide us with the tools to count selections where order is irrelevant. While permutations answered the question "How many ways can we arrange?" combinations answer the question "How many ways can we choose?" This distinction is crucial for solving counting problems accurately and, by extension, for calculating probabilities correctly in scenarios involving selections rather than arrangements.

Just as permutations enhanced our ability to solve probability problems involving arrangements, combinations will enable us to tackle probability questions involving selections. Many real-world probability scenarios involve choosing items from groups without regard to order:

• Selecting lottery numbers
• Choosing committee members
• Drawing cards for a poker hand
• Picking items for quality control testing
• Forming teams or groups

By mastering combination techniques, we'll be able to calculate probabilities for these selection-based scenarios accurately and efficiently.

In this section, we will learn to distinguish clearly between situations requiring permutations (where order matters) and those requiring combinations (where order doesn't matter). We'll develop the combination formula, explore its relationship to permutations, and apply combination techniques to solve both counting problems and probability questions. Most importantly, we'll build your intuition for recognizing when to use combinations versus permutations in practical problem-solving situations.

 

In the previous section, we saw an example that there were 12 ways to choose a President and a Vice-President from a group of 4 different people named Amy, Bill, Cassidy, and Daphne.

Now, let’s pose a different scenario: How many ways can you choose two secretary positions from a group of 4 different people? So, if we use the abbreviations of the previous four-different people from before, Amy, Bill, Cassidy, & Daphne, and we look at those 12 choices we obtained from the previous section, then we again have the following:

AB

BA

CA

DA

AC

BC

CB

DB

AD

BD

CD

DC

 

Observe that both AB and BA appear in our results. However, since we are selecting two secretaries without assigning specific roles (different from our President and Vice-President example where positions mattered), these represent the same selection. We are double-counting this outcome. Since ORDER DOES NOT MATTER in combinations, we must remove all such duplicate selections, which reduces our total count to six combinations:

AB

 

 

 

AC

BC

 

 

AD

BD

CD

 

 

So, what if the above problem had more people or more of a selection for the secretary position or both? This would take a great amount of time and effort to answer that question. Luckily, there is a formula that can be used to directly compute the total number of permutations without actually listing them as illustrated above.

Definition: Combinations

The number of ways to select \(r\) objects from a group of \(n\) objects, where the order of selection does not matter, is denoted by \(_nC_r\) and is given by the formula:

\[ _nC_r=\frac{n!}{(n-r)!\cdot r!} \nonumber \]

 

Example #5.4.1 🤔

How many ways can you choose two secretary positions from a group of \(4\) different people?

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✅ Solution:

So, ORDER DOES NOT MATTER here, because the secretary positions do not have ranks.

Thus, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 4\) and \(r = 2\).

\[ _4C_2=\frac{4!}{(4-2)!\cdot 2!}=\frac{4!}{2!\cdot 2!}=\frac{24}{2\cdot 2}=\frac{24}{4}=6 \nonumber \]

There are \(6\) different ways.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _4C_2=6 \nonumber \]

Example #5.4.2 🤔

In how many ways can a sorority of \(20\) members select three members to serve on a committee?

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✅ Solution:

So, ORDER DOES NOT MATTER here, because the sorority members do not have ranks.

Thus, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 20\) and \(r = 3\).

\[ _{20}C_3=\frac{20!}{(20-3)!\cdot 3!}=\frac{20!}{17!\cdot 3!}=\frac{20\cdot19\cdot18\cdot17!}{17!\cdot 6}=\frac{20\cdot19\cdot18\cdot \color{red}\cancel{\color{black}17!}}{\color{red}\cancel{\color{black}17!}{\color{black}\cdot 6}}=\frac{20\cdot19\cdot18}{6}=\text{1,140} \nonumber \]

There are \(\text{1,140}\) different ways.

Note: Observe that 20! was not expanded out entirely in our work. The reason is that 20! produces a number too large for standard calculators, which would display the result in scientific notation. Since \(20! = 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot \cdots \cdot 3 \cdot 2 \cdot 1\), we can rewrite it as \(20! = 20 \cdot 19 \cdot 18 \cdot 17!\). We expand only to \(17!\) because this will cancel directly with the \(17!\) in the denominator. The key strategy is to expand factorials just enough to create terms that will cancel with factorials in the denominator.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _{20}C_3=\text{1,140} \nonumber \]

Example #5.4.3 🤔

A textbook search committee is considering seven books for possible adoption. The committee has decided to select three of the seven for further consideration. In how many ways can they do so?

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✅ Solution:

So, ORDER DOES NOT MATTER here, because choosing the three books do not have any ranks to them.

Thus, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 7\) and \(r = 3\).

\[ _7C_3=\frac{7!}{(7-3)!\cdot 3!}=\frac{7!}{4!\cdot 3!}=\frac{7\cdot6\cdot5\cdot4!}{4!\cdot 6}=\frac{7\cdot6\cdot5\cdot \color{red}\cancel{\color{black}4!}}{\color{red}\cancel{\color{black}4!}{\color{black}\cdot 6}}=\frac{7\cdot6\cdot5}{6}=\text{35} \nonumber \]

There are \(35\) different ways.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _7C_3=35 \nonumber \]

Example #5.4.4 🤔

A hand of five-card draw poker consists of an unordered arrangement of five cards from a standard deck of \(52\) cards. How many five-card draw poker hands are possible?

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✅ Solution:

So, ORDER DOES NOT MATTER here. The order in which you receive your cards or arrange them in your hand does not matter because the hand's value is based solely on the combination of cards you hold, not the sequence in which they were dealt.

Thus, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 52\) and \(r = 5\).

\[ _{52}C_5=\frac{52!}{(52-5)!\cdot 5!}=\frac{52!}{47!\cdot 5!}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47!}{47!\cdot 120}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot \color{red}\cancel{\color{black}47!}}{\color{red}\cancel{\color{black}47!}{\color{black}\cdot 120}}=\frac{52\cdot51\cdot50\cdot49\cdot48}{120}=\frac{\text{311,875,200}}{120}=\text{2,598,960} \nonumber \]

There are \(\text{2,598,960}\) different ways.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _{52}C_5=\text{2,598,960} \nonumber \]

Example #5.4.5 🤔

In 1988, the California Lottery was initiated. There were 49 lotto balls, each numbered from 1 to 49. In order to play, you must choose 6 numbers. If you pick all 6 winnings numbers, you win the jackpot. In how many ways can someone choose 6 numbers from the 49?

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✅ Solution:

So, ORDER DOES NOT MATTER here. The order in which you choose your numbers does not matter, because the lottery is based solely on the combination of numbers you choose, not the sequence in which they they were chosen.

Thus, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 49\) and \(r = 6\).

\[ _{49}C_6=\frac{49!}{(49-6)!\cdot 6!}=\frac{49!}{43!\cdot 6!}=\frac{49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43!}{43!\cdot 720}=\frac{49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot \color{red}\cancel{\color{black}43!}}{\color{red}\cancel{\color{black}43!}{\color{black}\cdot 720}}=\frac{49\cdot48\cdot47\cdot46\cdot45\cdot44}{720}=\frac{\text{10,068,347,520}}{720}=\text{13,983,816} \nonumber \]

There are \(\text{13,983,816}\) different ways.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _{49}C_6=\text{13,983,816} \nonumber \]

Example #5.4.6 🤔

Refer to Example #5.4.5. In 1990, the California Lottery changed its main Lotto game from a 6/49 format to a 6/53 format to increase jackpots. The change created 53 lotto balls, each numbered from 1 to 53, choosing 6 numbers for each play. In how many ways can someone choose 6 numbers from the 53?

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✅ Solution:

So, ORDER DOES NOT MATTER here. The order in which you choose your numbers does not matter, because the lottery is based solely on the combination of numbers you choose, not the sequence in which they they were chosen.

Thus, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 53\) and \(r = 6\).

\[ _{53}C_6=\frac{53!}{(53-6)!\cdot 6!}=\frac{53!}{47!\cdot 6!}=\frac{53\cdot52\cdot51\cdot50\cdot49\cdot48\cdot47!}{47!\cdot 720}=\frac{53\cdot52\cdot51\cdot50\cdot49\cdot48\cdot \color{red}\cancel{\color{black}47!}}{\color{red}\cancel{\color{black}47!}{\color{black}\cdot 720}}=\frac{53\cdot52\cdot51\cdot50\cdot49\cdot48}{720}=\frac{\text{16,529,385,600}}{720}=\text{22,957,480} \nonumber \]

There are \(\text{22,957,480}\) different ways.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _{53}C_6=\text{22,957,480} \nonumber \]

Example #5.4.7 🤔

How many ways can \(4\) cars and \(5\) trucks be selected from \(6\) cars and \(8\) trucks to be tested for a safety inspection?

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✅ Solution:

So, ORDER DOES NOT MATTER here, because their is no mention of any ranks.

Now, in this situation there are are two parts. We need to select \(4\) cars from a total of \(6\) and then we need to select \(5\) trucks from a total of \(8\) trucks.

For the cars, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 6\) and \(r = 4\).

\[ _6C_4=\frac{6!}{(6-4)!\cdot 4!}=\frac{6!}{2!\cdot 4!}=\frac{6\cdot5\cdot4!}{2\cdot 4!}=\frac{6\cdot5\cdot \color{red}\cancel{\color{black}4!}}{\color{black}2\color{red}\cancel{\color{black}\cdot 4!}}=\frac{6\cdot5}{2}=\text{15} \nonumber \]

For the trucks, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 8\) and \(r = 5\).

\[ _8C_5=\frac{8!}{(8-5)!\cdot 5!}=\frac{8!}{3!\cdot 5!}=\frac{8\cdot7\cdot6\cdot5!}{5!\cdot 6}=\frac{8\cdot7\cdot6\cdot \color{red}\cancel{\color{black}5!}}{\color{red}\cancel{\color{black}5!}{\color{black}\cdot 6}}=\frac{8\cdot7\cdot6}{6}=\text{56} \nonumber \]

Since we need to select cars AND trucks, we multiply using the Fundamental Counting Principle from Section 5.2:

\[= _6C_4 \cdot  _8C_5 \nonumber \]

\[=15 \cdot 56 \nonumber \]

\[=840 \nonumber \]

There are \(840\) different ways.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _6C_4 \cdot  _8C_5=840 \nonumber \]

Example #5.4.8 🤔

There is a group of \(6\) men and \(8\) women. How many ways can you select two men and two women from that group?

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✅ Solution:

So, ORDER DOES NOT MATTER here, because their is no mention of any ranks.

Now, in this situation there are are two parts. We need to select \(2\) men from a group of \(6\) men and then we need to select \(2\) women from a group of \(8\) women.

For the men, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 6\) and \(r = 2\).

\[ _6C_2=\frac{6!}{(6-2)!\cdot 2!}=\frac{6!}{4!\cdot 2!}=\frac{6\cdot5\cdot4!}{4!\cdot 2}=\frac{6\cdot5\cdot \color{red}\cancel{\color{black}4!}}{\color{red}\cancel{\color{black}4!}{\color{black}\cdot 2}}=\frac{6\cdot5}{2}=\text{15} \nonumber \]

For the women, we will use the formula for combinations

\[\ _nC_r=\frac{n!}{(n-r)!\cdot r!}, \nonumber \]

where \(n = 8\) and \(r = 2\).

\[ _8C_2=\frac{8!}{(8-2)!\cdot 2!}=\frac{8!}{6!\cdot 2!}=\frac{8\cdot7\cdot6!}{6!\cdot 2}=\frac{8\cdot7\cdot \color{red}\cancel{\color{black}6!}}{\color{red}\cancel{\color{black}6!}{\color{black}\cdot 2}}=\frac{8\cdot7}{2}=\text{28} \nonumber \]

Since we need to select mens AND women, we multiply using the Fundamental Counting Principle from Section 5.2:

\[= _6C_2 \cdot  _8C_2 \nonumber \]

\[=15 \cdot 28 \nonumber \]

\[=420 \nonumber \]

There are \(420\) different ways.

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✅ Alternative Solution:

You can also use a calculator function to get directly to the solution

\[ _6C_2 \cdot  _8C_2=420 \nonumber \]

Section 5.4: Combinations [In-Class Exercises]

1. Out of 7 volunteers, a teacher must choose 2 to sit on a discussion panel. How many panels can be formed?

2. A student government has 12 members. They need to choose a 3‑person subcommittee. How many different 3‑person committees can be formed?

3. An ice cream shop offers 10 toppings, and you want 4 different toppings on a sundae. In how many ways can you choose 4 toppings?

4. A class has 18 students. The teacher randomly chooses 2 students to pair up for an example demonstration. How many possible student pairs are there?

5. A group has 9 movies available and wants to choose 3 to show during a weekend marathon. They don’t care about the order they’re shown, just which movies are chosen. How many sets of 3 movies can be selected?

6. A counselor wants to randomly choose 5 students from a group of 30 to participate in a survey interview. How many selections of 5 students are possible?

7. A grocery store display has 7 types of fruit and 5 types of vegetables. A shopper wants to pick 2 fruits and 3 vegetables. How many different produce selections are possible?.

8. At BJ's Restaurant & Brewhouse, you can customize your pizza with the following toppings.
   • Meat: Cupping Pepperoni, Italian Sausage, Smoked Bacon, Chicken, Smoked Ham, Housemade Meatballs, & Anchovies
   • Veggie: Mushrooms, Black Olives, Jalapenos, Pineapple, Green Bell Peppers, Fresh Basil, Roasted Garlic, Fire Roasted Red Peppers, Onions, Red Onions, & Green Onions

    If we were to get a pizza with three different meat toppings and three different vegetable toppings, how many different ways can you order a pizza?

9. The mathematics division at a local community college includes 5 calculus instructors, 4 statistics instructors, and 3 trigonometry instructors. The Vice President of Instruction wants to select 2 instructors from each subject area to speak at a high‑school recruitment event. How many different groups of instructors can be formed?

10. Challenge. In the game Bingo, a BINGO card has 5 rows and 5 columns, thus providing 25 total spaces. The columns are labeled from left  to right with the letters: 'B', 'I', 'N', 'G', 'O'. Each column contains 5 distinct numbers from 15 different numbers. With one exception, the center space in the 'N' column is a "free" space, so the "N" column will only contain 4 distinct numbers. So, each space in the 'B' column contains a number from 1 to 15, each space in the 'I' column contains a number from 16 to 30, each space in the 'N' column contains a number from 31 to 45, each space in the 'G' column contains a number from 46 to 60, and each space in the 'O' column contains a number from 61 to 75. How many different BINGO cards can be made under these configurations.

Answers

1) \(_7C_2=21\);   2) \(_{12}C_3=220\);   3) \(_{10}C_4=210\);   4) \(_{18}C_2=153\);   5) \(_9C_3=84\);   6) \(_{30}C_5=\text{142,506}\);  7) \(_7C_2\cdot_5C_3=210\);   8) \(_7C_3\cdot_{11}C_3=\text{5,775}\);   9) \(_5C_2\cdot_4C_2\cdot_3C_2=\text{1,200}\);   10) \(\left( 5!\cdot\left(_{15}C_{5}\right)\right)^{4}\cdot \left( 4!\cdot\left(_{15}C_{4}\right)\right)  = \text{552,446,474,061,128,648,601,600,000} \approx 5.52×10^{26}\).

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