Solution: 2017 Fall Final - 7

Author: Michiel Smid

Question

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

BOOKKEEPER

(a)

$2! \cdot 2! \cdot 3!$

(b)

${10 \choose 2} \cdot {8 \choose 2} \cdot {6 \choose 3} \cdot {3 \choose 2}$

(c)

${10 \choose 2} \cdot {8 \choose 2} \cdot {6 \choose 3} \cdot 3 \cdot 2$

(d)

${10 \choose 2} \cdot {8 \choose 2} \cdot {5 \choose 3} \cdot 3 \cdot 2$

First, we choose 2 of the 10 positions for the $O$‘s: $ \binom{10}{2} $

Then, we choose 2 of the remaining 8 positions for the $K$‘s: $ \binom{8}{2} $

Then, we choose 3 of the remaining 6 positions for the $E$‘s: $ \binom{6}{3} $

Then we choose 1 of the remaining 3 positions for the $B$: $ \binom{3}{1} $

Then we choose 1 of the remaining 2 positions for the $P$: $ \binom{2}{1} $

Then we choose 1 of the remaining 1 positions for the $R$: $ \binom{1}{1} $

Adding it up is $ \binom{10}{2} \cdot \binom{8}{2} \cdot \binom{6}{3} \cdot \binom{3}{1} \cdot \binom{2}{1} \cdot \binom{1}{1} $

Simplifying, we get $ \frac{10!}{2!} \cdot \frac{8!}{2!} \cdot \frac{6!}{3!} \cdot 3! $