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?

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.

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 .

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.

first, find the total combinations. let's start with using 2 republicans

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

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.

## Replies to: Hard Math Question

1,290Senior MemberThe 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.

1,581Senior Member3,187Senior Member513Member998MemberCombinations can easily be done without a calculator.

n!/[(r!)((n-r)!)]

3,187Senior Member240Junior Member123

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.

998MemberFor 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.

3,187Senior Member1,581Senior Member3,187Senior Member1,079Senior Member99Junior MemberXY = democrats

first, find the total combinations. let's start with using 2 republicans

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

1,581Senior Member3,187Senior MemberI 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.