Difference between revisions of "2011 AIME I Problems/Problem 12"
(→Solution) |
m (→Solution) |
||
Line 19: | Line 19: | ||
The first two cases gives us all the possible ways to arrange the people. Let there be <math>n</math> women. For the first case, if we think of (n) as dividers, we get <math>\dbinom{n+3}{3}</math> ways. For the second cases, we get <math>\dbinom{n+2}{2}</math> cases. | The first two cases gives us all the possible ways to arrange the people. Let there be <math>n</math> women. For the first case, if we think of (n) as dividers, we get <math>\dbinom{n+3}{3}</math> ways. For the second cases, we get <math>\dbinom{n+2}{2}</math> cases. | ||
− | The third to | + | The third to fifth cases counts the cases we desires. |
The third and fourth cases give us <math>2\dbinom{n+1}{2}</math> if we put 1 woman between (2) and (4) before we count. | The third and fourth cases give us <math>2\dbinom{n+1}{2}</math> if we put 1 woman between (2) and (4) before we count. | ||
Revision as of 07:14, 29 March 2011
Problem
Six men and some number of women stand in a line in random order. Let be the probability that a group of at least four men stand together in the line, given that everyman stands next to at least one other man. Find the least number of women in the line such that does not exceed 1 percent.
Solution
Denote (n) be n consecutive men and _ between (n) and (m) be some number of women between the mens(it can be zero).
There are five cases to consider:
_(2)_(2)_(2)_
_(3)_(3)_
_(2)_(4)_
_(4)_(2)_
_(6)_
The first two cases gives us all the possible ways to arrange the people. Let there be women. For the first case, if we think of (n) as dividers, we get ways. For the second cases, we get cases.
The third to fifth cases counts the cases we desires. The third and fourth cases give us if we put 1 woman between (2) and (4) before we count.
the last case gives us
so the probability is
the numerator simplifies into .
The denominator simplifies into
so the whole faction simplifies into
Since is slightly less than 1 when is large. will be close to . They equals to each other when .
If we let or , we will notices that the answer is
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |