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Hard Math Question

edited July 2012
13. There are 3 Republicans and 2 Democrats on a Senate committee. If a 3-person subcommittee is to be formed from this committee, what is the probability of selecting 2 Republicans and 1 Democrat?

(A) 1/20
(B) 3/20
(C) 3/10
(D) 3/5
(E) 2/3

It's (D). WHY.
edited July 2012
38 replies
Post edited by risubu on
«1

Replies to: Hard Math Question

• 1266 replies24 threads Senior Member
The total number of ways to form a 3 person subcommittee is 5C3 = 10 (you're choosing 3 people from 5).

The total number of ways to choose 2 republicans from 3 is 3C2 = 3.

The total number of ways to choose 1 democrat from 2 is 2C1 = 2.

By the counting principle, there are 3*2 = 6 ways to select 2 republicans and 1 democrat.

Thus the answer is 6/10 = 3/5.
• 1417 replies164 threads Senior Member
Thanks for the great explanation! You make it seem like a no-brainer
• 2907 replies281 threads Senior Member
Is there a way to do this problem without a calculator?
nCr = n!/(r!(n-r)!)
Is there a way to do this problem without a calculator?

Combinations can easily be done without a calculator.

n!/[(r!)((n-r)!)]
• 2907 replies281 threads Senior Member
I should have clarified my question. Would there have been a way to solve this problem without knowing the nCr (combination) formula?
• 192 replies48 threads Junior Member
You can make a quick and dirty list. Say the Rs are 1, 2, 3 and the Ds are 4, 5.

123
124
125
134
135
145
234
235
245
345

That's it. How many have 2 from {1, 2, 3} and 1 from {4, 5}?

124, 125, 134, 135, 234, 234 all work. That's 6 of the 10.

nCr is great, but it's nice to have this as an option too.
I view combinations as doing two permutations(simply counting).

For example, when finding the number of ways a subgroup of 2 can be formed from a group of 5, one can either:

use 5C2 = 10

or

find how many groups of 2 can be formed; 5*4 = 20
divide by the number of times a group is repeated; AB = BA;2
20/2 = 10

In order to find the number of times a group is repeated, one utilizes another counting principle by asking a question, "how many groups can be formed from 2 people?" = 2*1 = 2

Personally, understanding where combinations formula comes from helped me greatly. Utilizing nCr doesn't yield some number that I don't truly understand; it's just a shortcut for doing two permutations.
• 2907 replies281 threads Senior Member
Thank you Pwnthesat. I was thrown by this problem because I don't think I have ever seen an actual SAT math problem require explicit knowledge of the combination formula. Thus far, the counting principle has worked wonders on every single real SAT math problem .
• 1417 replies164 threads Senior Member
You're right this isn't a real SAT problem
• 2907 replies281 threads Senior Member
Where did you get this problem from? Was it the SAT Math Bible?
• 1059 replies20 threads Senior Member
The only time I have ever seen problems that "needed" combinatorics beyond the counting principle, the numbers involved were small enough to list. I'm thinking of that blue book problem about the plumber and the two apprentices. (Sounds like a joke -- but no, they do NOT walk into a bar..) So I say start by listing, switch to the counting principle when the numbers get big and leave nCr and nPr for the level II subject test.
• 94 replies5 threads Junior Member
ABC = republicans
XY = democrats

AB with X
AB with Y

AC with X
AC with Y

BC with X
BC with Y

now 2 democrats with 1 repub

XY with A
XY with B
XY with C

and finally, 3 republicans

ABC

the total is 10, and the question asks for the chance of 2 republicans and 1 democrat. just look back and we see there were 6 times where we could have 2 republicans with 1 democrat, so the answer is 6/10 or 3/5
• 1417 replies164 threads Senior Member
IceQube wrote:
Where did you get this problem from? Was it the SAT Math Bible?
No, chung's. Hence the stupidity.
• 2907 replies281 threads Senior Member
pckeller wrote:
The only time I have ever seen problems that "needed" combinatorics beyond the counting principle, the numbers involved were small enough to list. I'm thinking of that blue book problem about the plumber and the two apprentices.

I vividly recall that problem. Even that problem did not require combinatorics. Just use the counting principle on the two apprentices and divide by 2 because sending Tom and Bob the Plumber is the same as sending Bob and Tom.
• 106 replies2 threads Junior Member
Ok, a lot of you are making it really really complicated for most people who aren't geniuses. I'd just say there are 5 people, and you need to pick 3 specific ones... That makes 3/5, and it takes less than 10 seconds to think that through
Ok, a lot of you are making it really really complicated for most people who aren't geniuses. I'd just say there are 5 people, and you need to pick 3 specific ones... That makes 3/5, and it takes less than 10 seconds to think that through