Permutations with Repeated Letters

Question

This question is taken from A First Course in Probability (8e) by Ross.

How many different arrangements can be formed from the letters PEPPER?

I understand that there are $6!$ permutations of the letters when the repeated letters are distinguishable from each other. And that for each of these permutations, there are $(3!)(2!)$ permutations within the Ps and Es. This means that the $6$$!$ total permutations accounts for the $(3!)(2!)$ internal permutations. Then, the explanation in the text states that there are $\frac{6!}{(3!)(2!)} = 60$ possible letter arrangements of the letters PEPPER.

I don't understand this last part. I thought that since the internal permutations were accounted for the total possible letter arrangements would be the $1 - \frac{(3!)(2!)}{6!}$. Can someone please explain the logic behind the last part? Thank you.

Answer 1

Let's do it smaller with $PPE$. If the letters $P$ are given an index then there are $3!=6$ possibilities:

• $P_1P_2E$
• $P_2P_1E$
• $P_1EP_2$
• $P_2EP_1$
• $EP_1P_2$
• $EP_2P_1$

If the indices are taken away then $P_1P_2E$ and $P_2P_2E$ both become $PPE$. It appears that possibility $PPE$ has been counted $2!=2$ times. To repair this we must divide by $2!$ and get $3$ as answer. This agrees with the fact that there are $3$ possibilities:

• $PPE$
• $PEP$
• $EPP$

Likewise in $PEPPER$ every permutation is originally counted $3!2!$ times, so we must divide $6!$ by $3!2!$ to repair.

Answer 2

Suppose we want to form a permutation of $P_1E_1P_2P_3E_2R$.

We know there are $6!$ ways to do this, since the letters are all distinct, but we could also do this by

1) forming a permutation of $PEPPER$, which can be done in, say, x ways;

2) assigning subscripts to the P's, which can be done in $3!$ ways; and then

3) assigning subscripts to the E's, which can be done in $2!$ ways.

Therefore $6!=x(3!)(2!)\;,$ so $\displaystyle x=\frac{6!}{3!2!}$

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Here is an alternate way to do this:

1) Choose the places for the E's, which can be done in $\dbinom{6}{2}$ ways.

2) Next choose the place for the R, which can be done in $\dbinom{4}{1}=4$ ways.

The P's must go in the remaining places, so we get $\dbinom{6}{2}\cdot4$ permutations of PEPPER.

Answer 3

Main question as posted by the OP

How many different arrangements can be formed from the letters PEPPER?

Main Doubts:

1. 6! permutations of the letters when the repeated letters are distinguishable from each other
2. And that for each of these permutations, there are (3!)(2!) permutations within the Ps and Es
3. This means that the 6 ! total permutations accounts for the (3!)(2!) internal permutations. Then, the explanation in the text states that there are 6!(3!)(2!)=60
4. I don't understand this last part. I thought that since the internal permutations were accounted for the total possible letter arrangements would be the 1−(3!)(2!)6!

Answering all the points to clarify most of the doubts:

5. For unique 6 letters answer would be 6! or 6P6 (Permutation) or calculate using Combination (6c1 * 5c1 * 4c1 * 3c1 * 2c1 * 1c1), why arrangement is different from mere selection since order of selection is also important and that's we can deduce the same using Combination as shown above

6. 3! for P and 2! for E comes into the picture since order of selection mentioned above lose its significance, you cannot distinguish which P or which E thus there's a reduction in the total number of arrangements, which were there for the unique letters and that's why this division becomes important and thus the solution is 6!/(3! * 2!)

7. Now, why it can't be 1−(3!)(2!)6!, since we aren't looking for probability and anything that says number of ways or arrangements has to be >= 1

8. Now let's see how do we arrive at the final solution 6!/(3! * 2!), in the PEPPER, out of 6 letters there are 3P, 2E and 1R, how can we select 3 slots out of 6 for 3 P, 6C3, why not 6P3, that's because you can't arrange 3 P's in different ways, now remaining places are 3, so we can select two places for E as 3C2 and finally only 1 place remaining for R, so the solution would be 6C3 * 3C2 * 1C1 = 6!/(3!*3!) * 3!/(2!*1!) * 1! = 6!/3!*2!

9. Let's consider same thing other way we consider 2 E first, then 3 P and last R, 6C2 * 4C3 * 1C1 = 6!/(4!*2!) * 4!/(3!*1!) * 1! = 6!/3!*2!

10. Let's take another way, take R first then P then E, 6C1 * 5C3 * 2C2 = 6!/(5!*1!) * 5!/(3!*2!) * 2!/(2!*0!) = 6!/3!*2!

Therefore which ever way you choose the answer remains same and its easy to deduce