# 4.4.2: Permutations with Similar Elements

- Page ID
- 11507

Learning Objectives

In this section you will learn to

- Count the number of possible permutations when there are repeated items

In this section we will address the following problem.

- In how many different ways can the letters of the word MISSISSIPPI be arranged?

This is an example of Permutations with Similar Elements.

## Permutations with Similar Elements

Let us determine the number of distinguishable permutations of the letters ELEMENT.

Suppose we make all the letters different by labeling the letters as follows.

\[E_1LE_2ME_3NT \nonumber\]

Since all the letters are now different, there are 7! different permutations.

Let us now look at one such permutation, say

\[LE_1ME_2NE_3T \nonumber\]

Suppose we form new permutations from this arrangement by only moving the E's. Clearly, there are 3! or 6 such arrangements. We list them below.

\begin{aligned} &\mathrm{LE}_{1} \mathrm{ME}_{2} \mathrm{NE}_{3} \\

&\mathrm{LE}_{1} \mathrm{ME}_{3} \mathrm{NE}_{2} \\

&\mathrm{LE}_{2} \mathrm{ME}_{1} \mathrm{NE}_{3} \mathrm{T} \\

&\mathrm{LE}_{2} \mathrm{ME}_{3} \mathrm{NE}_{1} \mathrm{T} \\

&\mathrm{LE}_{3} \mathrm{ME}_{2} \mathrm{NE}_{1} \mathrm{T} \\

& \mathrm{LE}_{3} \mathrm{ME}_{I} \mathrm{NE}_{2} \mathrm{T} \end{aligned}

Because the E's are not different, there is only one arrangement LEMENET and not six. This is true for every permutation.

Let us suppose there are n different permutations of the letters ELEMENT.

Then there are \(n \cdot 3!\) permutations of the letters \(E_1LE_2ME_3NT\).

But we know there are 7! permutations of the letters \(E_1LE_2ME_3NT\).

Therefore, \(n \cdot 3! = 7!\)

Or \(n = \frac{7!}{3!}\).

This gives us the method we are looking for.

Definition: Permutations with Similar Elements

The number of permutations of n elements taken \(n\) at a time, with \(r_1\) elements of one kind, \(r_2\) elements of another kind, and so on, is

\[\frac{n !}{r_{1} ! r_{2} ! \ldots r_{k} !} \]

Example \(\PageIndex{3}\)

Find the number of different permutations of the letters of the word MISSISSIPPI.

**Solution**

The word MISSISSIPPI has 11 letters. If the letters were all different there would have been 11! different permutations. But MISSISSIPPI has 4 S's, 4 I's, and 2 P's that are alike.

So the answer is \(\frac{11!}{4!4!2!} = 34,650\).

Example \(\PageIndex{4}\)

If a coin is tossed six times, how many different outcomes consisting of 4 heads and 2 tails are there?

**Solution**

Again, we have permutations with similar elements.

We are looking for permutations for the letters HHHHTT.

The answer is \(\frac{6!}{4!2!} = 15\).

Example \(\PageIndex{5}\)

In how many different ways can 4 nickels, 3 dimes, and 2 quarters be arranged in a row?

**Solution**

Assuming that all nickels are similar, all dimes are similar, and all quarters are similar, we have permutations with similar elements. Therefore, the answer is

\[\frac{9 !}{4 ! 3 ! 2 !}=1260 \nonumber\]

Example \(\PageIndex{6}\)

A stock broker wants to assign 20 new clients equally to 4 of its salespeople. In how many different ways can this be done?

**Solution**

This means that each sales person gets 5 clients. The problem can be thought of as an ordered partitions problem. In that case, using the formula we get

\[\frac{20 !}{5 ! 5 ! 5 ! 5 !}=11,732,745,024 \nonumber\]

Example \(\PageIndex{7}\)

A shopping mall has a straight row of 5 flagpoles at its main entrance plaza. It has 3 identical green flags and 2 identical yellow flags. How many distinct arrangements of flags on the flagpoles are possible?

**Solution**

The problem can be thought of as distinct permutations of the letters GGGYY; that is arrangements of 5 letters, where 3 letters are similar, and the remaining 2 letters are similar:

\[ \frac{5 !}{3 ! 2 !} = 10 \nonumber\]

Just to provide a little more insight into the solution, we list all 10 distinct permutations:

GGGYY, GGYGY, GGYYG, GYGGY, GYGYG, GYYGG, YGGGY, YGGYG, YGYGY, YYGGG

We summarize.

Summary

**Permutations with Similar Elements**

The number of permutations of n elements taken n at a time, with \(r_1\) elements of one kind, \(r_2\) elements of another kind, and so on, such that \(\mathrm{n}=\mathrm{r}_{1}+\mathrm{r}_{2}+\ldots+\mathrm{r}_{\mathrm{k}}\) is

\[\frac{n !}{r_{1} ! r_{2} ! \dots r_{k} !} \nonumber\]

This is also referred to as **ordered partitions**.