Solution: 2014 Fall Midterm - 4

Author: Michiel Smid

Question

Consider 12 boys, 17 girls, and 25 dogs. How many ways are there to arrange them on a line if

all boys stand next to each other,
all girls stand next to each other,
all dogs stand next to each other,
none of the boys are standing next to any of the girls.

(a)

$(12!)(17!)(25!)$

(b)

$2(12!)(17!)(25!)$

(c)

$2(12! + 17! + 25!)$

(d)

$12! + 17! + 25!$

Solution

We can break it down

Let B be the # of ways to place boys: $|B| = 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 12!$

Let G be the # of ways to place girls: $|G| = 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 17!$

Let D be the # of ways to place girls: $|D| = 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 \cdot 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 25!$

Imagine (B,D), and (G) are all groups. Boy group can’t be beside the girl group, so the arrangement will either be (BDG) or (GDB).

That’s 2 possible 2 combos of groups

$2 \cdot |B| \cdot |G| \cdot |D| = 2 \cdot 12! \cdot 17! \cdot 25! = 2(12!)(17!)(25!)$