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C ( 5 , 0 ) + C ( 5 , 1 ) + C ( 5 , 2 ) + C ( 5 , 3 ) + C ( 5 , 4 ) + C ( 5 , 5 ) = 32

There are 32 possible pizzas. This result is equal to 2 5 .

We are presented with a sequence of choices. For each of the n objects we have two choices: include it in the subset or not. So for the whole subset we have made n choices, each with two options. So there are a total of 2 · 2 · 2 · · 2 possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time.

Formula for the number of subsets of a set

A set containing n distinct objects has 2 n subsets.

Finding the number of subsets of a set

A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?

We are looking for the number of subsets of a set with 4 objects. Substitute n = 4 into the formula.

2 n = 2 4      = 16

There are 16 possible ways to order a potato.

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A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?

64 sundaes

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Finding the number of permutations of n Non-distinct objects

We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be 12 ! ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the 12 ! permutations we counted are duplicates. The general formula for this situation is as follows.

n ! r 1 ! r 2 ! r k !

In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are 4 ! ways to order the stars and 3 ! ways to order the moon.

12 ! 4 ! 3 ! = 3 , 326 , 400

There are 3,326,400 ways to order the sheet of stickers.

Formula for finding the number of permutations of n Non-distinct objects

If there are n elements in a set and r 1 are alike, r 2 are alike, r 3 are alike, and so on through r k , the number of permutations can be found by

n ! r 1 ! r 2 ! r k !

Finding the number of permutations of n Non-distinct objects

Find the number of rearrangements of the letters in the word DISTINCT.

There are 8 letters. Both I and T are repeated 2 times. Substitute n = 8 ,   r 1 = 2 ,   and   r 2 = 2   into the formula.

8 ! 2 ! 2 ! = 10 , 080  

There are 10,080 arrangements.

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Find the number of rearrangements of the letters in the word CARRIER.

840

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Access these online resources for additional instruction and practice with combinations and permutations.

Key equations

number of permutations of n distinct objects taken r at a time P ( n , r ) = n ! ( n r ) !
number of combinations of n distinct objects taken r at a time C ( n , r ) = n ! r ! ( n r ) !
number of permutations of n non-distinct objects n ! r 1 ! r 2 ! r k !

Key concepts

  • If one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways. See [link] .
  • If one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m × n ways. See [link] .
  • A permutation is an ordering of n objects.
  • If we have a set of n objects and we want to choose r objects from the set in order, we write P ( n , r ) .
  • Permutation problems can be solved using the Multiplication Principle or the formula for P ( n , r ) . See [link] and [link] .
  • A selection of objects where the order does not matter is a combination.
  • Given n distinct objects, the number of ways to select r objects from the set is C ( n , r ) and can be found using a formula. See [link] .
  • A set containing n distinct objects has 2 n subsets. See [link] .
  • For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See [link] .

Questions & Answers

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Practice Key Terms 5

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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