Math question. My professor and me have different answer. Please help.

Question

A squad of 4 soldiers is selected from a 15 man platoon to keep watch each night.

How many different squad can be selected if either Bob or Roy ( but never both) must be on the squad watch all the time?

My professor said it's 2* 13C2

But I think it's 2* 13C3

Please correct me.

I really don't know about professor answer. But mine Explanation is Bob or roy must always be present in any team thus forming a team is: 13C3 x 2 = 572 Combination is used because order is not considered. and x 2 since the 13 men would be interchangeable but the 2 must be always there thus. Multiply 2 to the combination of the squad formed.

Answer 1

There is only one right answer - in this case, $572$. But as is often the case in combinatorics, there are many perfectly valid ways to arrive at the answer.

One way is to divide the two platoon into two groups:

$A = \{\mbox{Bob}, \mbox{Roy}\}$

$B = \{\mbox{everyone else}\}$

Choose one member from the first group, and three members from the second group. There are $2{13 \choose 3} = 572$ ways to do this, as you concluded.

Here's a totally different way:

1. Count all the possible squads. ${15 \choose 4} = 1365$.
2. Count all the squads that contain neither Bob nor Roy. ${13 \choose 4} = 715$. We reject these.
3. Count all the squads that contain both Bob and Roy. ${13 \choose 2} = 78$. We reject these too.
4. So the answer is ${15 \choose 4} - ({13 \choose 4} + {13 \choose 2}) = 1365 - (715 + 78) = 572$. This agrees with the earlier answer.

This generalizes to a combinatorial identity: $2 {n - 2 \choose k - 1} = {n \choose k} - \left( {n - 2 \choose k} + {n - 2 \choose k - 2} \right)$. It is possible to prove this by moving the negative terms on the right side over to the left side and then combining using the identity ${n \choose k} + {n \choose k + 1} = {n + 1 \choose k + 1}$ twice.