Arrangements with Duplicates

Problems involving arrangements with duplicates sometimes come up on the GRE. Consider the following question: How many different ways can you arrange the letters in the word _tarp_ in a sequence? How might you explain how to get the answer to the question?

Correct!

Incorrect. A combination of four elements taken four at a time is a collection of four elements in which order does not matter. It this case, order matters. The number of combinations of four objects taken four at a time is actually equal to one.

Incorrect. There are more than four ways to arrange the four letters in a sequence. Just to get you started, you have _tapr_, _ptar_, _arpt_, _rpta_, and _trap_.

Consider the following question: How many ways can the letters in the word _tree_ be arranged in a sequence? How does this question differ from the previous question?

Incorrect.

Consider the word _Mississippi_. How many items are duplicated?

Correct!

Incorrect.

Incorrect.

There are three items duplicated. The _i_'s are a group of duplicates with four identical duplicates. The _p_'s are a group of duplicates with two identical duplicates. The _s_'s are a group of duplicates with four identical duplicates. So, how many ways can you arrange the letters in the word _Mississippi_?

Incorrect. This would be true if there were no repeated letters. But the fact that there are repeated letters makes the situation rather different.

Incorrect.

Correct!

To summarize: [[summary]]

That's right!

The number of ways that four letters can be arranged in a sequence is equal to the number of permutations of four objects taken four at a time. "Arranging in a sequence" means that you place the letters in some particular order. Thus, _rapt_, _trap_, _rtpa_, and so forth. Consider the formula for ways to arrange a list of _k_ elements is _k_!. In this case there are 4! = 24 ways to arrange the letters in the word _tarp_.

Since the letter _e_ appears twice in the word _tree_, the situation is rather different. You must apply a different formula to this problem to count the numbers of arrangements. Here is how this situation is handled: If a list of _k_ items is arranged, there are _k_! ways to do that. For every item that occurs more than once, divide _k_! by the _x_!, where _x_ is the number of times that item occurs.

In the word _tree_, the letter _e_ is repeated twice, so there is one item that occurs more than once and it occurs twice. This means that the number of permutations is given by $$\displaystyle \frac{4!}{2!}=12$$.

To obtain the number of permutations when there are duplicates, for every item duplicated you divide by the number of duplicates factorial. In this case you have _k_ = 11, since you are interested in the arrangement of 11 letters. Since there is an item duplicated with two identical duplicates and two items duplicated with four identical duplicates, the number of permutations is equal to $$\displaystyle \frac{11!}{4!2!4!}=\frac{11!}{(24)(2)(24)}=34,650$$. So there are 34,650 ways to arrange the letters in the word _Mississippi_.

The answer is the number of permutations of four elements taken four at a time; in other words, it is equal to 4!.

The answer is the number of combinations of four elements taken four at a time; in other words, it is equal to $$\frac{4!}{4!}$$.

The answer is just equal to 4, since there are four letters.

It is the same situation, so the answer is clearly 24.

It is different because the letter _e_ is repeated twice.

Three

Four

Ten

$$11!$$

$$\frac{11!}{3!}$$

$$\frac{11!}{4!2!4!}$$

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