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Now we address the second problem.
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.
Since all the letters are now different, there are $7!$ different permutations.
Let us now look at one such permutation, say
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.
Because the $E$ 's are not different, there is only one arrangement $\text{LEMENET}$ and not six. This is true for every permutation.
Let us suppose there are $n$ different permutations of the letters $\text{ELEMENT}$ .
Then there are $n\cdot 3!$ permutations of the letters ${E}_{1}{\text{LE}}_{2}{\text{ME}}_{3}\text{NT}$ .
But we know there are $7!$ permutations of the letters ${E}_{1}{\text{LE}}_{2}{\text{ME}}_{3}\text{NT}$ .
Therefore, $n\cdot 3!=7!$
Or $n=\frac{7!}{3!}$ .
This gives us the method we are looking for.
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
Find the number of different permutations of the letters of the word MISSISSIPPI.
The word MISSISSIPPI has 11 letters. If the letters were all different there would have been $\text{11}!$ different permutations. But MISSISSIPPI has 4 S's, 4 I's, and 2 P's that are alike.
So the answer is $\frac{\text{11}!}{4!4!2!}$
Which equals 34,650.
If a coin is tossed six times, how many different outcomes consisting of 4 heads and 2 tails are there?
Again, we have permutations with similar elements.
We are looking for permutations for the letters HHHHTT.
The answer is $\frac{6!}{4!2!}=\text{15}$ .
In how many different ways can 4 nickels, 3 dimes, and 2 quarters be arranged in a row?
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
A stock broker wants to assign 20 new clients equally to 4 of its salespeople. In how many different ways can this be done?
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
We summarize.
The number of permutations of $n$ elements in a circle is
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 $n={r}_{1}+{r}_{2}+\cdots +{r}_{k}$ is
$\frac{n!}{{r}_{1}!{r}_{2}!\cdots {r}_{k}!}$
This is also referred to as ordered partitions .
Suppose we have a set of three letters $\left\{A,B,C\right\}$ , and we are asked to make two-letter word sequences. We have the following six permutations.
$\text{AB}$ $\text{BA}$ $\text{BC}$ $\text{CB}$ $\text{AC}$ $\text{CA}$
Now suppose we have a group of three people $\left\{A,B,C\right\}$ as Al, Bob, and Chris, respectively, and we are asked to form committees of two people each. This time we have only three committees, namely,
$\text{AB}$ $\text{BC}$ $\text{AC}$
When forming committees, the order is not important, because the committee that has Al and Bob is no different than the committee that has Bob and Al. As a result, we have only three committees and not six.
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