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Hard Math Question

risuburisubu 1417 replies164 threads Senior Member
edited July 2012 in SAT Preparation
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
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Replies to: Hard Math Question

  • DrSteveDrSteve 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.
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  • risuburisubu 1417 replies164 threads Senior Member
    Thanks for the great explanation! You make it seem like a no-brainer :)
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  • IceQubeIceQube 2907 replies281 threads Senior Member
    Is there a way to do this problem without a calculator?
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  • semaphore12semaphore12 488 replies25 threads Member
    nCr = n!/(r!(n-r)!)
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  • JefferyJungJefferyJung 974 replies24 threads Member
    Is there a way to do this problem without a calculator?

    Combinations can easily be done without a calculator.

    n!/[(r!)((n-r)!)]
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  • IceQubeIceQube 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?
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  • PWNtheSATPWNtheSAT 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. :)
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  • JefferyJungJefferyJung 974 replies24 threads Member
    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.
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  • IceQubeIceQube 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 :).
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  • risuburisubu 1417 replies164 threads Senior Member
    You're right this isn't a real SAT problem ;)
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  • IceQubeIceQube 2907 replies281 threads Senior Member
    Where did you get this problem from? Was it the SAT Math Bible?
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  • pckellerpckeller 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.
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  • emidankemidank 94 replies5 threads Junior Member
    ABC = republicans
    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
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  • risuburisubu 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.
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  • IceQubeIceQube 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.
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  • ClarkU2016ClarkU2016 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
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  • JefferyJungJefferyJung 974 replies24 threads 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

    Your logic is flawed. Perhaps you should think about your statement again.
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  • IceQubeIceQube 2907 replies281 threads Senior Member
    ^You tell him :)!
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  • ClarkU2016ClarkU2016 106 replies2 threads Junior Member
    Could you explain to me how it's flawed please
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  • PWNtheSATPWNtheSAT 192 replies48 threads Junior Member
    Well, what if the question asked for 2 democrats and 1 republican instead? Would you still get 3/5?
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