Solution: 2014 Fall Midterm - 3

Author: Michiel Smid

Question

Consider 17 boys and 17 girls. How many ways are there to arrange them on a line if all boys are standing next to each other and all girls are standing next to each other?

(a)

$17! + 17!$

(b)

$2(17!)^2$

(c)

$2(17! + 17!)$

(d)

$(17!)^2$

Solution

We can break it down

Let B be the # of ways to place boys

$|B| = 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 17!$

Let G be the # of ways to place girls

$|G| = 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 17!$

Now, all the boys are together and all the girls are together.

We can rearrange these groups in 2 ways: Boys on the left and girls on the right OR Boys on right right and girls on the left.

These are two possible ways to rearrange these groups. As such, we multiply by 2:

$2 \cdot |B| \cdot |G| = 2 \cdot 17! \cdot 17! = 2{(17!)}^2$