Difference between revisions of "2001 AMC 12 Problems/Problem 16"
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| − | You can put all <math>8</math> socks on first for <math>8!</math> ways and then all <math>8</math> shoes on next for <math>8!</math> more ways. This is not the only possibility, so the lower bound is <math>(8!)^2</math>. You can choose all <math>16</math> in a random fashion, but some combinations would violate the rules, so the upper bound is <math>16!</math>. <math>\text{C}</math> & <math>\text{E}</math> are the lower and upper bounds, so the answer is in between them, <math>\boxed{\frac {16!}{2^8}}</math>. | + | You can put all <math>8</math> socks on first for <math>8!</math> ways and then all <math>8</math> shoes on next for <math>8!</math> more ways. This is not the only possibility, so the lower bound is <math>(8!)^2</math>. You can choose all <math>16</math> in a random fashion, but some combinations would violate the rules, so the upper bound is <math>16!</math>. <math>\text{(C)}</math> & <math>\text{(E)}</math> are the lower and upper bounds, so the answer is in between them, <math>\boxed{\frac {16!}{2^8}}</math>. |
- anandiyer12 | - anandiyer12 | ||
Revision as of 21:24, 3 March 2016
Problem
A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?
Solution
Solution 1
Let the spider try to put on all 
things in a random order. Each of the 
permutations is equally probable. For any fixed leg, the probability that he will first put on the sock and only then the shoe is clearly 
. Then the probability that he will correctly put things on all legs is 
. Therefore the number of correct permutations must be 
.
Solution 2
Each dressing sequence can be uniquely described by a sequence containing two 
s, two 
s, ..., and two 
s -- the first occurrence of number 
means that the spider puts the sock onto leg , the second occurrence of means he puts the shoe onto leg .
If the numbers were all unique, the answer would be . However, since 8 terms appear twice, the answer is 
.
Solution 3
You can put all socks on first for 
ways and then all shoes on next for more ways. This is not the only possibility, so the lower bound is 
. You can choose all in a random fashion, but some combinations would violate the rules, so the upper bound is . 
& 
are the lower and upper bounds, so the answer is in between them, .
- anandiyer12
See Also
| 2001 AMC 12 (Problems • Answer Key • Resources) | |
| Preceded by Problem 15 |
Followed by Problem 17 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
