Permutations with Repeating Letters

Solution 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.

Solution 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 $\displaystyle x=\frac{6!}{3!2!}$

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.