Question 472664: Dear math teacher,
I really need help with the following word problem:
There are 4 hooks on a wall. In how many ways can 3 coats be hung on them, one coat on a hook?
I solved it several times and was kind of sure of an answer but my book says 24, and I got 12. Here is how I did it:
4 times Permutation of taking 3 coats and hanging them one at a time because we have 4 hooks. That gives me 12 ways. I also drew a picture of coat 1, coat 2, and coat 3, and started hanging each coat on hook 1, 2, 3, and 4 making a tree below each coat. Each tree gave me 4 ways to hand a coat on a hook, and for 3 coats, I simply added 4 + 4 + 4 ways = 12 ways. So, I got the same answer twice. But then, I did permutation of taking 4 hooks and hanging 3 coats on them and I got 16 ways but this does not make sense because units to the left of permutation must match the units to the right of permuation and I have 4 hooks to the left and 3 coats to the right, so I knew that's is the wrong approach. This kind of reassured me that the first two approaches are correct.
Please help me figure out this problem. I would really appreciate it. Thank you so much. And have a wonderful day.
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! nPr=permutation of n things taken r at a time.=n!/(n-r)!
4!/(4-3)!=4!=24
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Ed
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123.
12.3
1.23
.123
132.
13.2
1.32
.132
213.
21.3
2.13
.213
231.
23.1
2.31
.231
312.
31.2
3.12
.312
321.
32.1
3.21
.321
.
The decimal point is the empty hook in each set of 4 numbers. It could be replace by 0 if you like. As you can see there are 24 ways just as the formula predicts.
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Ed
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