0.10 Sets and counting  (Page 7/9)

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.

We summarize.

Combinations

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.