# Thread: Numbers & Operations Q from SAT Sbj Math II Test Study

1. ## Numbers & Operations Q from SAT Sbj Math II Test Study

I'm sorry if this is in the wrong sub-forum. I'm guessing this could be classified as pretty basic stat. But here's the problem:

"From a group of 6 juniors and 8 seniors on the student council, 2 juniors and 4 seniors will be chosen to make up a 6-person committee. How many different 6-person committees are possible?

a. 84
b. 85
c. 1050
d. 1710
e. 1890
"

Then it tells me:
"Choice (c) is the correct answer. The 2 juniors on the committee can be chosen from the 6 juniors in (6/2) = 15 ways. The 4 seniors on the committee can be chosen from the 8 seniors in (8/4) = 70 ways. Therefore there are (15)(70) = 1050 possibilities for the 6 person committee."

I'm pretty bad at these.. Can someone please explain (6/2) = 15 and (8/4) = 70 deal to me? Or maybe just show me a clearer way of doing this? Thanks!

2. Hello, Savior_Self!

rom a group of 6 juniors and 8 seniors on the student council,
2 juniors and 4 seniors will be chosen to make up a 6-person committee.
How many different 6-person committees are possible?

. . $(a)\;84 \qquad (b)\; 85 \qquad (c)\;1050 \qquad (d)\; 1710 \qquad (e)\; 1890$

Then it tells me: ."Choice (c) is the correct answer.
The 2 juniors on the committee can be chosen from the 6 juniors in (6/2) = 15 ways.
The 4 seniors on the committee can be chosen from the 8 seniors in (8/4) = 70 ways.
Therefore, there are (15)(70) = 1050 possibilities for the 6 person committee."

Can someone please explain (6/2) = 15 and (8/4) = 70 deal to me?

Those expressions are not fractions; they are "combination" numbers.

Selecting 2 juniors from 6 juniors, there are: $\displaystyle {6\choose2} \:=\:\frac{6!}{2!\,4!} \:=\:15$ ways.

Selecting 4 seniors from 8 seniors, there are: $\displaystyle {8\choose4} = \frac{8!}{4!\,4!} \:=\:70$ ways.

Maybe you'd recognize them if they were written like this?

. . . . . $_6C_2\:\text{ and }\:_8C_4$