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Gmat prep Combination

This topic has 1 expert reply and 5 member replies
vladmire Master | Next Rank: 500 Posts Default Avatar
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Gmat prep Combination

Post Mon Dec 08, 2008 5:15 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    The lunch menu at a certain restaurant contains 4 different entrees and 5 different side dishes. If a meal consists of 1 entree and 2 different side dishes, how many different meal combinations could be chosen from this menu?

    10
    20
    40
    80
    100

    I know I have to use the combinations formula nCr but cannot figure which numbers go where.

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    pbanavara Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Dec 08, 2008 5:35 pm
    I guess it's 80 - 4c1*5p2

    The first entree can be chosen in 4c1 ways
    The two side dishes can be chosen in 5p2 ways because they have to be different - you cannot choose the same dish twice. If this limitation was not there then it would be 5c2 I guess ..

    I'll let others chime in - P&C is not my cup of T anyways

    - pradeep

    niraj_a Legendary Member Default Avatar
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    Post Mon Dec 08, 2008 5:54 pm
    i would say 40

    4C1 * 5C2 = 4 * 10 = 40

    Thanked by: pragga
    cramya Legendary Member Default Avatar
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    Post Mon Dec 08, 2008 7:59 pm
    Its a combination prob since same 1 entree same 2 dish is the same meal no matter u pick the entree first or the 2 dishes first

    40

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    Stuart Kovinsky GMAT Instructor
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    Post Mon Dec 08, 2008 10:05 pm
    pbanavara wrote:
    I guess it's 80 - 4c1*5p2

    The first entree can be chosen in 4c1 ways
    The two side dishes can be chosen in 5p2 ways because they have to be different - you cannot choose the same dish twice. If this limitation was not there then it would be 5c2 I guess ..

    I'll let others chime in - P&C is not my cup of T anyways

    - pradeep
    We use permutations when order matters and combinations when it doesn't.

    Let's say our 5 side dishes are salad, rice, fries, corn and jelly beans. Do we care if we pick salad then corn or corn then salad? No, we end up with the same set of dishes.

    Since order doesn't matter, we use combinations.

    The combinations formula is:

    nCk = n!/k!(n-k)!

    in which n = total # of objects and k = # chosen.

    In this question, since we're choosing entrees AND side dishes, we apply the forumla twice and then multiply the results. If we were choosing entrees OR side dishes, we'd apply the formula twice and then add the results.

    So, as noted above, we have:

    4C1 * 5C2

    4!/1!3! * 5!/2!3!

    4*3*2*1/3*2*1 * 5*4*3*2*1/2*1*3*2*1
    4 * 10
    40

    (Note that I wrote out the entire factorial, but there are lots of short cuts for cancelling out and, on test day, we certainly don't need to write down all those 1s, since they don't have any impact on our calculations.)

    Thanked by: pbanavara, vladmire, Baldini
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    vladmire Master | Next Rank: 500 Posts Default Avatar
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    Post Tue Dec 09, 2008 5:19 pm
    What's a good example of when order does matter for use of the Permutation formula N!/(n-r)!

    anayeri Senior | Next Rank: 100 Posts Default Avatar
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    Post Tue Dec 09, 2008 9:03 pm
    Say you have to pick pairs of letters from the three letters A,B,C.

    - If order matters, then AB and BA are two different pairs (ie it matters whether A occupies the first slot or the second). So we have 3p2 = 3!/(3-1!) = 6 possibilities {AB, BA, AC, CA, BC, CB}

    - If order doesn't matter, then AB and BA are the same pair. So we have 3c2=3!/(2!*(3-2)!) = 3 possibilities {AB/BA, AC/CA, BC/CB}.

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