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risubu
1417 replies164 threads Senior Member

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.

38 replies(A) 1/20

(B) 3/20

(C) 3/10

(D) 3/5

(E) 2/3

It's (D). WHY.

Post edited by risubu on

This discussion has been closed.

## Replies to: Hard Math Question

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.

Combinations can easily be done without a calculator.

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

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.

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.

XY = 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

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.

Your logic is flawed. Perhaps you should think about your statement again.