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Combination Problem

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Combination Problem Post Wed Nov 12, 2014 12:31 pm
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    There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

    (A)15
    (B)16
    (C)28
    (D)56
    (E)64

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    Post Wed Nov 12, 2014 12:36 pm
    datonman wrote:
    There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

    (A)15
    (B)16
    (C)28
    (D)56
    (E)64
    There are 8 teams. If we ask each team, "How many teams did you play?" we'll find that each team played 7 teams, which gives us a total of 56 games (since 8 x 7 = 56).

    From here we need to recognize that each game has been COUNTED TWICE.
    For example, if Team A and Team B play a game, then Team A counts it as a game, and Team B ALSO counts it as a game.

    So, to account for the DUPLICATION, we'll divide 56 by 2 to get 28 (C)
    --------------------------
    Here are three related questions:
    http://www.beatthegmat.com/ugghhh-i-pick-this-question-from-btg-forum-t267675.html
    http://www.beatthegmat.com/number-of-handshakes-t100109.html
    http://www.beatthegmat.com/soccer-time-t70860.html

    Cheers,
    Brent

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    Post Wed Nov 12, 2014 12:36 pm
    datonman wrote:
    There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

    (A)15
    (B)16
    (C)28
    (D)56
    (E)64
    Each game must be played by a UNIQUE PAIR of teams.
    Thus, the total number of games = the total number of unique pairs that can be formed from the 8 teams.
    Number of options for the first team selected = 8. (Any of the 8 teams.)
    Number of options for the second team selected = 7. (Any of the 7 remaining teams.)
    To combine these options, we multiply:
    8*7.
    Since the ORDER of the two teams doesn't matter -- AB is the same pair as BA -- we divide by the number of ways the two teams can be arranged (2!):
    (8*7)/(2*1) = 28.

    The correct answer is C.

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    Post Wed Nov 12, 2014 12:56 pm
    datonman wrote:
    There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

    (A)15
    (B)16
    (C)28
    (D)56
    (E)64
    As noted above, each game must be played by a UNIQUE PAIR of teams.
    An alternate approach is to WRITE OUT all of the possible pairings.
    Let the 8 teams be A, B, C, D, E, F, G and H.

    Pairs with A:
    AB, AC, AD, AE, AF, AG, AH.
    7 options.

    Pairs with B (excluding AB, which has already been counted):
    BC, BD, BE, BF, BG, BH.
    6 options.

    Pairs with C (excluding AC and BC, which have already been counted):
    CD, CE, CF, CG, CH.
    5 options.

    Notice the PATTERN.
    Options for A = 7.
    Options for B = 6.
    Options for C = 5.
    With each successive letter, the number of options decreases by 1.
    Thus:
    Options for D = 4.
    Options for E = 3.
    Options for F = 2.
    Options for G = 1.

    Adding together the options above, we get:
    Total number of games = 7+6+5+4+3+2+1 = 28.

    The correct answer is C.

    _________________
    Mitch Hunt
    GMAT Private Tutor
    GMATGuruNY@gmail.com
    If you find one of my posts helpful, please take a moment to click on the "Thank" icon.
    Available for tutoring in NYC and long-distance.
    For more information, please email me at GMATGuruNY@gmail.com.

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    Post Fri Nov 14, 2014 6:18 am
    Another way to solve this quickly is to sketch 8 dots in a circle and then connect each dot in turn to every other dot. There will be 28 lines. Answer = C.

    Post Fri Nov 14, 2014 6:41 am
    datonman wrote:
    There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played?

    (A)15
    (B)16
    (C)28
    (D)56
    (E)64
    We can also solve this question using COMBINATIONS.

    It takes two teams to play a game.
    So, in how many different ways can we select 2 teams from 8 teams?
    Since the order in which we select the teams doesn't matter, we can use combinations.
    We can select 2 teams from 8 teams in 8C2 ways
    8C2 = 28

    If anyone is interested, we have a free video on calculating combinations (like 8C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

    Cheers,
    Brent

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    Post Sun Nov 16, 2014 8:58 am
    Somehow I have known the answer practically since last 22 years. 1992 Cricket world cup was played on all play all basis and my father explained it to me with the approach of A plays B the B plays A, but I wanted to know in terms of Combination.

    gautamkumar Rising GMAT Star Default Avatar
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    Post Sun Nov 16, 2014 8:59 am
    Somehow I have known the answer practically since last 22 years. 1992 Cricket world cup was played on all play all basis and my father explained it to me with the approach of A plays B the B plays A, but I wanted to know in terms of Combination.

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    Post Wed Nov 19, 2014 7:06 am
    Another quick solution is to consider the 8 teams as 8 integers used in a 2 digit number.
    In base 8, the highest value is 77 (base 8) = 7x8 + 8 = 64 (i.e there are 64 combinations of pairs of 8 integers)
    Then we need to exclude double digits (8 of them) to get 56
    and halve the answer to remove palindromic values (AB = BA), to get 56/2 = 28
    Job done!

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