KA-0172Selections of a few objects
4 points, difficulty 2 of 3Level 11–12about 120sA committee of 4 must be formed from 5 boys and 4 girls, and it must contain exactly 2 boys and 2 girls. How many different committees are possible?
Hints
Take them one at a time. The first gives nothing away.
1A nudge
Choose the boys and the girls as two separate jobs.
2The strategy
The number of ways to pick 2 boys from 5 is 5 choose 2, and 2 girls from 4 is 4 choose 2.
3The full solution
5 choose 2 = 10 and 4 choose 2 = 6, and the two choices are independent, so 10 x 6 = 60.
Solution
The reliable way
Split the task into two independent choices. Picking 2 boys from 5 can be done in C(5,2) = (5 x 4)/2 = 10 ways. Picking 2 girls from 4 can be done in C(4,2) = (4 x 3)/2 = 6 ways. Every choice of boys can pair with every choice of girls, so the multiplication principle gives 10 x 6 = 60 committees. The transferable idea: independent choices multiply, and only mutually exclusive cases add.
The elegant way
C(5,2) x C(4,2) = 10 x 6 = 60.
Why this is on the test: Adding when you should multiply, and ignoring a composition constraint, are the two errors that account for most lost marks here.