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A family has 4 girls and 3 bedrooms. 2 of the bedrooms are only big enough 1 girl, and the last room is big enough for 2 girls. How many ways are there to assign the girls to the bedrooms?

I came up with 4!/2! I thought that because there are 4 girls, there are 4! ways of putting them into the rooms. With 2 of the rooms being able to house 1 girl, and 1 room being able to house 2 girls, I thought it would be 4!/1!1!2!. Is this correct?

Tyler
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    The number is correct. There should be some explanation of the reasoning, You could do it in a more basic way. Suppose the two small rooms are painted red and blue. There are $4$ ways to choose who will be in the red room, and for each choice there are $3$ ways to choose who will be in the blue room. This determines the full room assignment, so the number is $(4)(3)$. – André Nicolas May 19 '15 at 06:21
  • @AndréNicolas You mean how I got my answer? Does my reasoning not make sense? – Tyler May 19 '15 at 06:22
  • It would be acceptable, but on the terse side. – André Nicolas May 19 '15 at 06:26

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True! There are 4 different ways to place a girl to the first small room, there are three ways to place a girl into the second small room and there is only one way to place remaining girls into the big room. Thus, there are 12 different ways to assign girls to rooms.

How many ways are there to assign girls to rooms assuming that each room can accommodate two girls?

wdacda
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  • 2 ways? If you put 2 girls into one room, then you have only two girls left for the remaining rooms? – Tyler May 19 '15 at 06:25
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    @Tyler There are $\binom{4}{2} = 6$ ways to select two of the girls to be placed in the first room. – N. F. Taussig May 19 '15 at 11:39
  • @N.F.Taussig Oh right because there are 4 girls and 2 girls can be chosen to be in each room. I hadn't thought of it like that. Thanks! – Tyler May 19 '15 at 18:39