Solution: 2014 Winter Final - 7

Author: Michiel Smid

Question

Consider 4 blue balls $B_1,B_2,B_3,B_4$ and 5 red balls $R_1,R_2,R_3,R_4,R_5$. We pick 3 balls of the same color and arrange them on a horizontal line. (The order on the line matters.) How many arrangements are there?

(a)

(b)

(c)

(d)

Solution

Let’s break it down

Let A = 3 blue balls
We choose 3 of the 4 blue balls $ \binom{4}{3} $
The first red ball has 3 possible positions, second has 2 possible remaining positions, and the third has 1 remaining possible position: $ 3! $
$ |A| = \binom{4}{3} \times 3! $
$ |A| = 4 \times 3! $
$ |A| = 24 $
Let B = 3 red balls
We choose 3 of the 5 red balls: $ \binom{5}{3} $
The first blue ball has 3 possible positions, second has 2 possible remaining positions, and the third has 1 remaining possible position: $ 3! $
$ |B| = \binom{5}{3} \times 3! $
$ |B| = 10 \times 3! $
$ |B| = 60 $

The total number of arrangements is the sum of the arrangements in sets $ A \cup B $:

$ |A| + |B| $

$ = 24 + 60 $

$ = 84 $