Solution: 2014 Fall Final - 7

Author: Michiel Smid

Question

Consider $m \geq 7$ blue balls $B_1,B_2,\dots,B_m$ and $n \geq 7$ red balls $R_1,R_2,\dots,R_n$. We pick 7 balls of the same color and arrange them on a horizontal line. (The order on the line matters.) How many arrangements are there?

(a)

$7!{m + n \choose 7}$

(b)

None of the above.

(c)

$7!{m \choose 7} + 7!{n \choose 7}$

(d)

$m!{m \choose 7} + n!{n \choose 7}$

Solution

A = Event that we pick 7 blue balls

First, we choose 7 of the m blue balls: $ \binom{m}{7} $

The first ball has 7 choices, the second ball has 6 choices, the third ball has 5 choices, and so on until the seventh ball has 1 choice: $ 7! $

$ |A| = \binom{m}{7} \cdot 7! $

B = Event that we pick 7 red balls

First, we choose 7 of the n red balls: $ \binom{n}{7} $

The first ball has 7 choices, the second ball has 6 choices, the third ball has 5 choices, and so on until the seventh ball has 1 choice: $ 7! $

$ |B| = \binom{n}{7} \cdot 7! $

$ |A| + |B| = \binom{m}{7} \cdot 7! + \binom{n}{7} \cdot 7! $