And finally, we want to address the case when we have permutations with repeated objects. What if we have something like:

Example
The DMV is wondering how many new license plates it can form from just the letters: T, T, and R. How many such license plates are there?

Now, this question is something of a hybrid between our permutations and combinations. It is true that the order of the objects matters: TTR is not the same license plate as RTT. But we can’t just treat it as a standard permutation where we calculate n!/(n-k)!. If we tried that, we would get:

3!/(3-3)! = 3! = 6

But, if we list out all the possibilities, we find that we cannot form 6 different possibilities. The only possible license plates are:

TTR

TRT

RTT

You can try to think of another ordering, but there isn’t one. So treating it as a normal permutation leads us to the wrong answer. What is going on here?

To see what’s going wrong, let’s label our two T’s as T1 and T2. Now, let’s see how many ways we can order them:

T1 T2 R

T1 R T2

T2 T1 R

T2 R T1

R T1 T2

R T2 T1

Now we get the six possibilities we were expecting. But, remember, as a license plate, it doesn’t matter whether one uses T1 or T2. At the end of the day, the letter that shows up on the plate is just T. So, for example, combinations like R T1 T2 are just the same as R T2 T1. They both result in license plates that look like:

R T T

So, like in our combination problem, our permutation is over-counting things. It is treating as distinct some possibilities that are, in fact, the same. Here, our permutation gives us 6 possible outcomes. But, in fact, the ones with the same color are the same:

T1 T2 R

T1 R T2

T2 T1 R

T2 R T1

R T1 T2

R T2 T1

And we see that, in fact, we only have 3 different possibilities, which is what we found in listing out the possible license plates.

So our rule for these cases:

Permutations with Repeated Objects
Suppose you want to order k objects where one of the objects repeats n times. Then, the number of possible orderings is: \frac{k!}{n!}.

is again a way of correcting for our over-counting. In our case, we can think of this rule as saying: there are 6 permutations. But within those permutations, the order of the T1 and T2 doesn’t matter. There are 2! ways to order T1 and T2, so the total number of distinct ways to order R, T, and T is just: 6/2! = 3.

Here is an application of this rule that involves larger numbers:

Example
You are playing Scrabble. Your hand has: T, R, S, S, S, E, and F. How many possible ways are there to order all of your tiles?

Solution
Following our rule, we want to find a way to order 7 objects (i.e. our 7 letters) where one of the objects is repeated 3 times. Then, there are 7!/3! ways to do this. And again,

7! = 7*6*5*4*3!

So,

7!/3! = 7*6*5*4 = 210

There are 210 possibilities

Practice Problems

Question 1
You are trying to come up with anagrams. You get the letters: A, Z, P, P, E, D. How many 6-letter combinations can be formed?

Answer

Question 2
You are playing Scrabble. You get the letters: S, S, S, S, T, E, P, X. How many 8-letter combinations can be formed?

Answer

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