Question

A defensive football squad consists of 25 players. Of these, 10 are linemen, 10 are linebackers, and 5 are safeties. How many different teams of 5 linemen, 3 linebackers and 3 safeties can be formed?

Short answer

$302400$ways

Step-by-step solution

. Given Information

We are given a football team of 25 players having 10 linemen,10 linebackers and 5 safeties. We need to find the number of ways to make a team of 5 linemen 3 linebackers and 3 safeties

. Finding the number of ways 

We need to choose 5 linemen from 10 linemen and 3 linebackers from 10 linebackers and 3 safeties from 5 safeties so the numbers of ways is :

$C(10,5)\times C(10,3)\times C(5,3)=\frac{10!}{5!5!}\times \frac{10!}{3!7!}\times \frac{5!}{3!2!}$

$\frac{10\times 9\times 8\times 7\times 6\times 5!}{5\times 4\times 3\times 2\times 1\times 5!}\times \frac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}\times \frac{5\times 4\times 3!}{2\times 1\times 3!}=\frac{10\times 9\times 8\times 7\times 6}{5\times 4\times 3\times 2}\times \frac{10\times 9\times 8}{3\times 2}\times \frac{5\times 4}{2}$

$3\times 2\times 7\times 6\times 5\times 3\times 8\times 5\times 2=302400$ways