Solution: 2013 Fall Midterm - 10

Author: Pat Morin

Question

How many strings can be obtained by rearranging the letters of the word POOPERSCOOPER

(a)

$13!$

(b)

$4!3!2!2!1!1!$

(c)

${{13}\choose{4}}{{9}\choose{3}}{{6}\choose{2}}{{4}\choose{2}}{{2}\choose{1}}$

(d)

${{13}\choose{4}}{{9}\choose{3}}{{6}\choose{2}}{{4}\choose{2}}$

Solution

There are 13 letters in the word “POOPERSCOOPER”.

We can break down how many letters of each there are:

3 P’s, 4 O’s, 2 E’s, 2 R’s, 1 S, 1 C

We choose 4 spots for the O's out of the 13 spots: $ \binom{13}{4} $
We choose 3 spots for the P's out of the 9 spots: $ \binom{9}{3} $
We choose 2 spots for the E's out of the 6 spots: $ \binom{6}{2} $
We choose 2 spots for the R's out of the 4 spots: $ \binom{4}{2} $
We choose 1 spot for the S out of the 2 spots: $ \binom{2}{1} $
We choose 1 spot for the C out of the 1 spot: $ \binom{1}{1} $

Thus, there are $ \binom{13}{4} \cdot \binom{9}{3} \cdot \binom{6}{2} \cdot \binom{4}{2} \cdot \binom{2}{1} \cdot \binom{1}{1} $ = $ \binom{13}{4} \cdot \binom{9}{3} \cdot \binom{6}{2} \cdot \binom{4}{2} \cdot \binom{2}{1} $ strings