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Permutations and Combinations: Practice Question of the Week #13

Here’s this week’s question– we’ll be posting up the answer tomorrow, good luck!

In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?

  1. 20
  2. 40
  3. 216
  4. 720
  5. 729


Update: Here’s the answer/explanation post!

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6 Responses to Permutations and Combinations: Practice Question of the Week #13

  1. nagababu June 30, 2014 at 2:57 pm #

    correct answer is A,20

  2. Siva August 24, 2011 at 9:43 am #

    yep… answer is 20, however this is a problem from Permutations and Combinations.

    • Margarette Jung
      Margarette August 24, 2011 at 2:10 pm #

      Hi, Siva

      You’re right– about both the answer and the mistake. Thanks for the heads up!

  3. madhuri August 24, 2011 at 2:11 am #

    the ans is 720… the first child can receive the shirt in 6 ways the 2nd boy in 5 ways and so on…. 6*5*4*3*2*1=720 ways

  4. Karthik August 23, 2011 at 12:05 pm #

    This one’s an easy one. (Rather a very basic question)

    Correct Answer: A) 20

    6! / (3!x3!) = 20

  5. Janmesh August 23, 2011 at 11:27 am #

    The correct answer is A. 20

    In the given example , there are two set of shirts (green and red) , each set consisting of the 3 identical shirt .
    Hence , the total no of ways in which shirts will be distributed amongst 6 children will be [ 6! / (3! * 3!) ] = 20

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