permutation vs combination problem

[super112]

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If a house has 4 rooms (A,B,C, and D) and you have 12 colors at your disposal, how many different ways can you paint the rooms if each is to be a different color?

I thought to do combination with no repetition because I thought order doesn't matter

 

[FeralisExtremum]

The simplest way is to just realize it's 12 x 10 x 9 x 8 outcomes intuitively - there are 12 colors you could potentially use for the first room, then 11 colors for the second (because they're all different colored rooms you can't re-use the same color twice), 9 for the third, and a remaining 8 for the fourth.

In terms of formulas, you would use permutation, not combination [so 12!/8! not 12!/(4!8!)] because in this case order does matter - painting the first room blue and the second room red is a different outcome than painting the first room red and the second room blue, for example.

 

[super112]

Thanks! You're right, it does make sense when you think about it intuitively. I also appreciate your explanation for the permutation, very helpful!

 

[100%permwinner]

Another way to look at is if you are counting items that are distinct from one another (i.e. not counting the same item more than once), permutation is appropriate here. Suppose you count the same items more than once (i.e. order doesn't matter), combination is applicable. The number of combinations is usually less the number of permutations because combination involves repetition.

 

[alexis2010]

If you have DAT destroyer look up question #86 on the QR section, it's very similar to it. As everybody said it's a permutation.