Question 601074: How many ways can you rearrange 12 eggs ? Do you use 12! Or permutations ?
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
1 way. Presumably one egg is just like the other, at least for most practical purposes. Hence, the number of arrangements of a dozen eggs must be the same as the number of arrangements of the letters AAAAAAAAAAAA, namely 1. If they are different, as in a dozen differently decorated easter eggs, then yes, 12! arrangements.
John
My calculator said it, I believe it, that settles it
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website! There are 12 places to pick a place for the 1st egg,
For each of those 12 places, there remain 11 places to put the 2nd egg.
That 12·11 ways to place the first two 3ggs.
Factoriala ARE permutations.
The best way is to think it out:
There are 12 ways to place egg #1 in the carton.
For each of those 12 ways to place the 1st egg, there
remain 11 places in the carton to place egg #2.
That's 12·11 ways to place the first 2 eggs in the
carton.
For each of those 12·11 ways to place the 1st 2 eggs,
there remain 10 places in the carton to place egg #3.
That's 12·11·10 ways to place the first 3 eggs in the
carton.
For each of those 12·11·10 ways to place the 1st 3 eggs,
there remain 9 places in the carton to place egg #4.
That's 12·11·10·9 ways to place the first 4 eggs in the
carton.
For each of those 12·11·10·9 ways to place the 1st 4
eggs, there remain 8 places in the carton to place egg
#5. That's 12·11·10·9·8 ways to place the first 5 eggs in
the carton.
For each of those 12·11·10·9·8 ways to place the 1st 5
eggs, there remain 7 places in the carton to place egg
#6. That's 12·11·10·9·8·7 ways to place the first 6 eggs
in the carton.
For each of those 12·11·10·9·8·7 ways to place the 1st 6
eggs, there remain 6 places in the carton to place egg
#7. That's 12·11·10·9·8·7·6 ways to place the first 7
eggs in the carton.
For each of those 12·11·10·9·8·7·6 ways to place the 1st
7 eggs, there remain 5 places in the carton to place egg
#8. That's 12·11·10·9·8·7·6·5 ways to place the first 8
eggs in the carton.
For each of those 12·11·10·9·8·7·6·5 ways to place the
1st 8 eggs, there remain 4 places in the carton to place
egg #9. That's 12·11·10·9·8·7·6·5·4 ways to place the
first 9 eggs in the carton.
For each of those 12·11·10·9·8·7·6·5·4 ways to place the
1st 9 eggs, there remain 3 places in the carton to place
egg #10. That's 12·11·10·9·8·7·6·5·4·3 ways to place the
first 10 eggs in the carton.
For each of those 12·11·10·9·8·7·6·5·4·3 ways to place
the 1st 10 eggs, there remain 2 places in the carton to
place egg #11. That's 12·11·10·9·8·7·6·5·4·3·2 ways to
place the first 11 eggs in the carton.
For each of those 12·11·10·9·8·7·6·5·4·3·2 ways to place
the 1st 11 eggs, there remain only 1 place in the carton
to place egg #12. That's 12·11·10·9·8·7·6·5·4·3·2·1 ways
to place all 12 eggs in the carton.
Answer: 12·11·10·9·8·7·6·5·4·3·2·1 = 12!
That's also the number of permutations of 12 things taken all 12
at a time.
nPn or P(n,n) = n!
Edwin
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