Expert: Josh - 3/16/2006

Question
An NFL coach sometimes uses a defense that utilizes 5 defensive linemen, 3 linebackers, and 3 in the defensive secondary (cornerbacks and deep backs). If his roster includes 7 defensive linemen, 6 linebackers, and 8 defensive backs, in how many different ways can he pick the 11 players to implement that particular defense?

i dont understand if i should add all the numbers and do a factorial of them

Answer
Hi Kevin,

This question may be generalized as a
"Pick k from K", then
"Pick m from M", then
"Pick n from N" combinatorial problem.

For brevity, let k=5 and K=7 be the required number and the total number of "defensive linemen", respectively.
Similarly, let m=3 and M=6 be the required number and the total number of "linebackers", respectively.
Finally, let n=3 and N=8 be the required number and the total number of "defensive backs", respectively.

The main thing to note is that members from each category are selected INDEPENDENTLY. That is, once k are selected from K, we move on to selecting m from a "separate" group of M (which has nothing to do with the previous k selections). Subsequently, n are selected from N, once again, the choices made here are NOT conditioned upon previous selections.

Under these circumstances, we MULTIPLY together the combinations corresponding to each of the three selection processes. Specifically, each selection process is understood to be a "pick a from A" problem.....[#1]

I presume that you have learnt about how combinations work. The number of combinations possible when n members are selected from a group of N is given by:
C(n,N)=N!/[n!(N-n)!] .....[#2]
where factorial n!=n*(n-1)*(n-2)*...*2*1 and 0!=1 by definition.

Note: The difference between combination and permutation is that the order of appearance does not matter in the former. In combination problems, we are simply interested in forming groups. Whether person A is selected ahead of person B or vice versa makes no difference.

Applying [#1] and [#2], the answer to the question is
C(k,K)*C(m,M)*C(n,N)
=C(5,7)*C(3,6)*C(3,8)
=(7!/[5!2!]) * (6!/[3!3!]) * (8!/[3!5!]) ...[#]
=(7*6/2) * (6*5*4/3!) * (8*7*6/3!)
=21 * 20 * 56 ...since 3!=6

[#] using the fact that 7! = 1*2*3*4*5*6*7 = 5!*6*7 etc.