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Permutations with Repetition

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Permutations of Subsets and Permutations with Repetition

Twelve violin students are competing for three seats in the orchestra: 1st Chair, 2nd Chair, and 3rd Chair. How many ways can a 1st Chair, 2nd Chair, and 3rd Chair be selected from the 12 competing students?

Guidance

Sometimes we want to order a select subset of a group. For example, suppose we go to an ice cream shop that offers 15 flavors. If we want to layer 3 scoops of different flavors on an ice cream cone, how many arrangements are possible? Here, the order matters so a chocolate, strawberry and vanilla cone is different than a strawberry, vanilla and chocolate cone. This is an example of permutations of a subset. We don’t need to know how many ways we can order all 15 flavors, just three choices. You have actually solved problems like this already using the Fundamental Counting Principle. There are 15 choices for scoop one, 14 choices for scoop two and 13 choices for scoop 3, so 15\times14\times13=2730 .

We can use factorials to solve this as well. Consider the expression: \frac{15\times14\times13\times12!}{12!}=\frac{15!}{(15-3)!}=\frac{n!}{(n-r)!} , where n represents the total number of elements in the set and r represents the number of elements in the subset we are selecting. Mathematically, this can be written using the notation _{15}P_3 or _nP_r . To summarize, if we wish to find the number of permutations of r elements selected from a larger set containing n elements, we can use the formula: _nP_r=\frac{n!}{(n-r)!} .

^* Note that some textbooks use the notation P^n_r to represent _nP_r .

We can evaluate this expression easily on the calculator as well. First, enter the value of n(15) , then go to MATH \rightarrow PRB, select 2: _nP_r . Now enter the value of r (3) to get the expression 15 _nP_r 3 on your screen. Press ENTER one more time and the result is 2730.

Example A

How many ways can a President, Vice President, Secretary and Treasurer be selecting from a club with ten members?

Solution: In the selection process here, the order matters so we are calculating the number of permutations of a subset of 4 members of the 10 member club. So,

_{10}P_4=\frac{10!}{(10-4)!}=\frac{10\times9\times8\times7\times{\color{red}6!}}{{\color{red}6!}}=10\times9\times8\times7=5040.

Example B

Consider the word, VIRGINIA. How many unique ways can these letters be arranged?

Solution: There are eight letters which can be arranged 8! ways. However, some of these arrangements will not be unique because there are multiple I’s in the word VIRGINIA. For example, if we let the three I’s be different colors, then we can see that there are several indistinguishable ways the I’s can be arranged.

\text{V{\color{red}I}RG{\color{blue}I}N{\color{green}I}A}, \ \text{V{\color{red}I}RG{\color{green}I}N{\color{blue}I}A}, \ \text{V{\color{blue}I}RG{\color{red}I}N{\color{green}I}A}, \ \text{V{\color{blue}I}RG{\color{green}I}N{\color{red}I}A}, \ \text{V{\color{green}I}RG{\color{red}I}N{\color{blue}I}A}, \ \text{V{\color{green}I}RG{\color{blue}I}N{\color{red}I}A}

In fact, there are 3! or 6 ways that the three I’s can be arranged that are indistinguishable when the arrangement of the remaining letters is constant. To figure out the number of unique arrangements of the 8 letters with 3 that are indistinguishable we can find the permutations of the 8 letters and divide by the permutations of the indistinguishable items.

\frac{8!}{3!}=\frac{8\times7\times6\times5\times4\times{\color{red}3!}}{{\color{red}3!}}=8\times7\times6\times5\times4=6720

Example C

Consider the word PEPPERS. How many unique arrangements can be made of these letters?

Solution: There are seven letters in total, three of which are P and two of which are E. We can expand upon what we did in the last example and divide by the number of ways each of these letters can be arranged.

\frac{7!}{3!2!}=\frac{7\times6\times5\times4\times{\color{red}3!}}{{\color{red}3!}\times2\times1}=\frac{7\times6\times5\times(2\times{\color{red}\cancel{2}})}{{\color{red}\cancel{2}}}=7\times6\times5\times2=420

Intro Problem Revisit In the orchestra selection process, the order matters so we are calculating the number of permutations of a subset of 3 members of the 12 member club. So,

_{12}P_3=\frac{12!}{(12-3)!}=\frac{12\times11\times10\times{\color{red}9!}}{{\color{red}9!}}=12\times11\times10=1320

Therefore, the orchestra slots can be filled in 1320 ways.

More Guidance

We can generalize the rule used in Examples B and C as follows. In a set with n elements, in which n_1, n_2, n_3, \ldots are indistinguishable, we can find the number of unique permutations of the n elements using the formula:

\frac{n!}{n_1!\times n_2!\times n_3!\times\ldots}

Guided Practice

1. Find _{10}P_6

2. On a team of 12 players, how many ways can the coach select players to receive on of each of the following awards (one award per player): most valuable player, best sportsmanship, most improved player.

3. How many ways can 5 yellow, 4 red and 3 green balls be arranged in a row?

Answers

1. _{10}P_6=\frac{10!}{(10-6)!}=\frac{10\times9\times8\times7\times6\times5\times{\color{red}4!}}{{\color{red}4!}}=151,200

2. _{12}P_3=\frac{12!}{(12-3)!}=\frac{12\times11\times10\times{\color{red}9!}}{{\color{red}9!}}=1,320

3. \frac{12!}{5!\times4!\times3!}=\frac{12\times11\times10\times9\times8\times7\times6\times{\color{red}5!}}{{\color{red}4!}\times4!\times3!}=\frac{{\color{red}12}\times11\times10\times9\times(4\times{\color{green}2})7\times{\color{blue}6}}{{\color{red}(4\times3)}\times{\color{green}2}\times{\color{blue}3\times2}}=11\times10\times9\times4\times7=27,720

Practice

Evaluate the following expressions.

  1. _8P_5
  2. _{11}P_8
  3. Evaluate and explain the results of each of the following: _5P_5, \ _5P_0, \ _5P_1
  4. Sarah needs to go to five different stores. How many ways can she go to two of them before lunch?
  5. In a race there are eleven competitors in a particular age group. How many possible arrangements are there for the top five finishers in this age group?
  6. How many ways can eight distinct raffle prizes be awarded to fifteen ticket holders.
  7. In a class of 24 students, there are six groups of four students. How many ways can a teacher select one group for each of three classroom maintenance responsibilities?
  8. At a birthday party there are 6 unique prizes to be given randomly to the 6 quests such that no one guest receives more than one prize. How many ways can this be done?
  9. How many unique ways can the letters in MISSISSIPPI be arranged?
  10. How many ways can two Geometry books, eight Algebra books and three Pre-Calculus books be arranged on a shelf if all the books of each respective subject are identical.
  11. At a math department luncheon, the department chair has three $20 gift certificates to the local coffee shop, five $25 gift certificates to a local bookstore and two state of the art calculators to give as prizes amongst the 10 department members. If each teacher receives one prize, how many unique distributions of prizes can be made?

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