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Math Help - [SOLVED] Permutation and Combination Question 3

  1. #1
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    [SOLVED] Permutation and Combination Question 3

    11 players are to be selected for a football game In the squad are 3 goal keepers, 6 defenders, 5 mid-fielders and 6 strikers. In how many ways can a coach pick his team if he needs 1 goal keeper, 4 defenders, 3 mid-fielders and 3 strikers?

    I get 3C1 * 6C4 * 5C3 *6C3 and get 9000.
    Am I right?

    In how many ways can 3 red, 1 black, 2 white and 2 blue balls be selected from 6 red, 2 black, 5 white and 4 blue balls; and in how many ways can they be arranged?

    I get 6C3 * 2C1 * 5C2 * 4C2
    =2,400

    For second part of question, I take 8!/(3!2!2!) = 1,680 But the answer sheet shows another answer. Am I wrong, or the answer sheet is wrong?
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  2. #2
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    Hello, azuki!

    11 players are to be selected for a football game.
    In the squad are 3 goal keepers, 6 defenders, 5 mid-fielders and 6 strikers.
    In how many ways can a coach pick his team
    if he needs 1 goal keeper, 4 defenders, 3 mid-fielders and 3 strikers?

    I get: . \left(_3C_1\right)\left(_6C_4\right)\left(_5C_3\ri  ght)\left(_6C_3\right) \:=\:9000

    Am I right? . Yes!


    In how many ways can 3 red, 1 black, 2 white and 2 blue balls
    be selected from 6 red, 2 black, 5 white and 4 blue balls.
    And in how many ways can they be arranged?

    I get: . \left(_6C_3\right)\left(_2C_1\right)\left(_5C_2\ri  ght)\left(_4C_2\right) \:=\:2,\!400 . Right!

    For second part, I take: . \frac{8!}{3!\,2!\,2!}\: =\: 1,\!680

    But the answer sheet shows another answer. .
    What is it?

    What was the original wording of the problem?

    "Arranged" is a very sloppy term.

    . . \begin{array}{c}\text{They can be arranged in a row.} \\<br />
\text{They can be placed in a circle.} \\<br />
\text{They can be stacked to form a }2\!\times\!2\!\times\!2\text{ cube.} \\<br />
\vdots \end{array}

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