Counting Theory - Discrete Math

Question

There was a question on my exam which asked something along the lines of:

"How many ways are there to order the letters in 'PEPPERCORN' if all the letters are used?"

Would this be 10! or would I need to do Permutation? What is the solution?

Answer 1

The word PEPPERCORN contains ten letters, we have ten positions to fill. Arrangements are distinguished by the positions of the letters. Notice that switching the positions of the Es within an arrangement does not produce an arrangement that is distinguishable from the original arrangement.

Here is a strategy for counting distinguishable arrangements:

1. Choose three of those ten positions for the three Ps.
2. Choose two of the remaining seven positions for the two Es.
3. Choose two of the remaining five positions for the two Rs.
4. Choose one of the remaining three positions for the C.
5. Choose one of the remaining two positions for the O.
6. Fill the remaining position with the N.

After you multiply the binomial coefficients you obtain and simplify, you will obtain a multinomial coefficient. The numbers in the denominator represent the number of ways the letters of the word PEPPERCORN can be permuted within an arrangement without producing an arrangement that is distinguishable from the original arrangement.

Answer 2

If all the letters were different, the answer would be $10!$.

So... in classic style, let's pretend they're all different. $P_1E_1P_2P_3E_2R_1COR_2N$. Then the answer is indeed $10!$ but then we need to do something about all the rearrangements of letters that were identical but we have subscripted.

So for each anagram above, there will be others that are identical except for the order of $P_1P_2P_3$. So we can divide through by the count of those variations (that occur across all the $10!$ options generated) and similarly for $E_1E_2$ and $R_1R_2$.

And for a pre-packaged system of understanding this, check up on multinomial coefficients.