Solution: 2017 Winter Midterm - 2

Author: Michiel Smid

Question

Let $b \geq 1$ and $g \geq 1$ be integers. Consider $b$ boys and $g$ girls. How many ways are there to arrange these kids on a line such that the leftmost kid is a girl?

(a)

$(b + g)! / b$

(b)

$(b + g)!$

(c)

$g \cdot (b + g - 1)!$

(d)

None of the above.

Solution

There are $ g $ ways to choose the leftmost kid: $ g $

We can take any of the $ b+g-1 $ kids to be the second kid: $ b+g-1 $

We can take any of the $ b+g-2 $ kids to be the third kid: $ b+g-2 $

This decrements until we reach the last kid: $ 1 $

Thus, there are $ g \cdot (b+g-1)! $ ways to arrange the kids.