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Combinations

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Combinations

You've been given a menu at a restaurant and you're trying to decide what to eat. You have a choice of 15 different entrees, 12 different side dishes and 4 different breads. How many possibilities do you have for lunch? Does it matter which order you choose them from the menu? In other words, when you order will it make any difference if you tell the waitress your bread order before your entree order?

Watch This

First watch this video to learn about combinations.

CK-12 Foundation: Chapter2CombinationsA

Then watch this video to see some examples.

CK-12 Foundation: Chapter2CombinationsB

Watch this video for more help.

Khan Academy Combinations

Guidance

If you think about the lottery, you choose a group of lucky numbers in hopes of winning millions of dollars. When the numbers are drawn, the order in which they are drawn does not have to be the same order as on your lottery ticket. The numbers drawn simply have to be on your lottery ticket in order for you to win. You can imagine how many possible combinations of numbers exist, which is why your odds of winning are so small!

Combinations are arrangements of objects without regard to order and without repetition, selected from a distinct number of objects. A combination of n objects taken r at a time ({_n}C_r) can be calculated using the formula:

{_n}C_r = \frac{n!}{r!(n-r)!}

Example A

Evaluate: {_7}C_2 .

{_7}C_2 &= \frac{7!}{2!(7-2)!}\\{_7}C_2 &= \frac{7!}{2!(5)!}\\  {_7}C_2 &= \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(5 \times 4 \times 3 \times 2 \times 1)}\\{_7}C_2 &= \frac{5,040}{(2)(120)}\\  {_7}C_2 &= \frac{5,040}{240}\\  {_7}C_2 &= 21

Example B

In how many ways can 3 desserts be chosen in any order from a menu of 10?

There are 10 menu items (n = 10) , and you are choosing 3 desserts in any order (r = 3) .

{_{10}}C_3 &= \frac{10!}{3!(10-3)!}\\{_{10}}C_3 &= \frac{10!}{3!(7)!}\\{_{10}}C_3 &= \frac{10 \times 9 \times 8}{3 \times 2 \times 1}\\{_{10}}C_3 &= 120

Example C

There are 12 boys and 14 girls in Mrs. Cameron's math class. Find the number of ways Mrs. Cameron can select a team of 3 students from the class to work on a group project. The team must consist of 2 girls and 1 boy.

There are groups of both boys and girls to consider. From the 14 girls (n = 14) in the class, we are choosing 2 (r = 2) .

\text{Girls:}\\{_{14}}C_2 &= \frac{14!}{2!(14 -2)!}\\{_{14}}C_2 &= \frac{14!}{2!(12)!}\\{_{14}}C_2 &= \frac{87,178,291,200}{2(479,001,600)}\\{_{14}}C_2 &= \frac{87,178,291,200}{958,003,200}\\{_{14}}C_2 &= 91

From the 12 boys (n = 12) in the class, we are choosing 1 (r = 1) .

\text{Boys:}\\{_{12}}C_1 &= \frac{12!}{1!(12-1)!}\\{_{12}}C_1 &= \frac{12!}{1!(11)!}\\{_{12}}C_1 &= \frac{479,001,600}{1(39,916,800)}\\{_{12}}C_1 &= \frac{479,001,600}{39,916,800}\\{_{12}}C_1 &= 12

Therefore, the number of ways Mrs. Cameron can select a team of 3 students (2 girls and 1 boy) from the class of 26 students to work on a group project is:

\text{Total combinations} = {_{14}}C_2 \times {_{12}}C_1=91 \times 12=1,092

Points to Consider

  • How does a permutation differ from a combination?

Guided Practice

There are 18 Democrats and 20 Republicans in a committee. Find the number of ways the committee can form a sub-committee consisting of 3 Democrats and 4 Republicans.

Answer:

There are groups of both Democrats and Republicans to consider. From the 18 Democrats (n = 18) in the committee, we are choosing 3 (r = 3) .

\text{Democrats:}\\{_{18}}C_3 &= \frac{18!}{3!(18 -3)!}\\{_{18}}C_3 &= \frac{18!}{3!(15)!}\\{_{18}}C_3 &= \frac{18 \times 17 \times 16}{3 \times 2 \times 1}\\{_{18}}C_3 &= \frac{4,896}{6}\\{_{18}}C_3 &= 816

From the 20 Republicans (n = 20) in the committee, we are choosing 4 (r = 4) .

\text{Republicans:}\\{_{20}}C_4 &= \frac{20!}{4!(20-4)!}\\{_{20}}C_4 &= \frac{20!}{4!(16)!}\\{_{20}}C_4 &= \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1}\\{_{20}}C_4 &= \frac{116,280}{24}\\{_{20}}C_4 &= 4,845

Therefore, the number of ways the committee can form a sub-committee consisting of 3 Democrats and 4 Republicans is:

\text{Total combinations} = {_{18}}C_3 \times {_{20}}C_4=816 \times 4,845=3,953,520

Practice

  1. Determine whether the following situations would require calculating a combination:
    1. Selecting 3 students to attend a conference in Washington, D.C.
    2. Selecting a lead and an understudy for a school play
    3. Assigning students to their seats on the first day of school
  2. In how many ways can you select 17 songs from a mix CD of a possible 38 songs?
  3. If an ice cream dessert can have 2 toppings, and there are 9 available, how many different selections can you make?
  4. If there are 17 randomly placed dots on a circle, how many lines can be formed using any 2 dots?
  5. A committee of 4 is to be formed from a group of 13 people. How many different committees can be formed?
  6. There are 4 kinds of meat and 10 veggies available to make wraps at the school cafeteria. How many possible wraps have 1 kind of meat and 3 veggies?
  7. There are 15 freshmen and 30 seniors in the Senior Math Club. The club is to send 4 representatives to the State Math Championships.
    1. How many different ways are there to select a group of 4 students to attend the State Math Championships?
    2. If the members of the club decide to send 2 freshmen and 2 seniors, how many different groupings are possible?

  8. Julia is going on vacation, and she wants to pack 5 t-shirts and 4 pairs of shorts in her suitcase. If Julia has 22 t-shirts and 9 pairs of shorts from which to chose in her closet, in how many possible ways can she pack t-shirts and shorts for her vacation?
  9. Of Major League Baseball's 16 National League teams and 14 American League teams, 4 teams from each league make the playoffs. How many possible groups of playoff teams are there in Major League Baseball?
  10. If Rick randomly chooses 3 months of the year to give up junk food, what is the probability that he gives up junk food during the first 3 months or the last 3 months of the year?

Vocabulary

combination

combination

The number of possible arrangements (_nC_r) of objects (r) without regard to order and without repetition selected from a distinct number of objects (n).

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