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Double check my math please.

“There are four men with one coat each, How many combos can be made without them getting the RIGHT coat, And without repeating combinations…”

So that means automatically, 4 possibles are knocked off, Also for my final answer doing it by hand. And pairing them up one by one, I came to the conclusion there is only 7 possibles… I did this.. V The people. The coats come after the people’s names. 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3

So the first four numbers going down represent the people, Then the numbers opposite them in the row equal the person’s coat they are holding… I kept going and managed only 7… ( I just gave you three of the possibles… So.. :D )

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Answer #1

I hate probability…

I’m assuming the “Right” coat is the coat they started off with. There’s 12 combos. Pretty sure. And here’s my reasoning.

The four men are #1-4. The four coats are a-d.

They start off 1a, 2b, 3c, 4d. But those combos don’t count because they are the “Right” coats. So

1b, 1c, 1d 2a, 2c, 2d 3a, 3b, 3d 4a, 4b, 4c

12 combos.

I could be completely wrong though, but it makes sense to me.

Answer #2

There are 24 possible combos, 1 of which they all get the right coat, 6 in which two of them score, and 8 where only one gets his coat. Coats are numbered 1-4 and the columns below correspond to the men - column 1 (man #1) the first 6 lines are excluded because man #1 gets his coat - etc. A dash appears after each combo that does NOT give ANY man his own coat — 9 combos.

1 2 3 4 1 2 4 3 1 3 2 4 1 3 4 2 1 4 2 3 1 4 3 2 2 1 3 4 2 1 4 3– 2 3 1 4 2 3 4 1– 2 4 1 3– 2 4 3 1 3 1 2 4 3 1 4 2– 3 2 1 4 3 2 4 1 3 4 1 2– 3 4 2 1– 4 1 2 3– 4 1 3 2 4 2 1 3 4 2 3 1 4 3 1 2– 4 3 2 1–

I know there has to be a more elegant way of showing this, but this ‘bean counting’ is all I can do right now.

Answer #3

;) Correct, ask your mom for farther help though, it looks good to me!

Answer #4

I’m almost certain its 16…

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