Difference between revisions of "2008 AIME II Problems/Problem 12"
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+ | === Solution 4 (Vandermonde's) === | ||
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+ | Let the number of blue flags on the first flag be <math>b</math> and define <math>g</math> similarly. Now note the bound <math>g \le b+1.</math> Now we cannot have any two consecutive greens. This condition is the same as "there are <math>b+1</math> spaces between each blue flag. How many ways can we put the <math>g</math> green flags in the <math>b+1</math> spaces?" Therefore given <math>b</math> blue flags on the first flag and <math>g</math> green flags on the first flag we have a total of <math>\displaystyle\binom{b+1}{g}.</math> Similarly there are <math>\displaystyle\binom{11-b}{9-g}</math> ways for the second flag. Summing over all possible possibilites we see the sum is <math>\sum_{g = 0}^{b+1} \sum_{n=0}^{10} \displaystyle\binom{b+1}{g}\displaystyle\binom{11-b}{9-g}.</math> Swapping the sum we have <math>\sum_{g = 0}^{b+1} \sum_{n=0}^{10} \displaystyle\binom{b+1}{g}\displaystyle\binom{11-b}{9-g} = \sum_{n=0}^{10} \sum_{g = 0}^{b+1} \displaystyle\binom{b+1}{g}\displaystyle\binom{11-b}{9-g}.</math> Then applying vandermondes yields <math> \sum_{n=0}^{10} \displaystyle\binom{12}{9}</math> so the sum is <math>11 \cdot \displaystyle\binom{12}{9}.</math> However, this answer overcounts. We cannot have no flags on a pole, so we subtract <math>2 \cdot \displaystyle\binom{11}{9}</math> since we can have empty flags for the first or second flag. Therefore the answer is <math>11 \cdot \displaystyle\binom{12}{9} - 2 \cdot \displaystyle\binom{11}{9} \pmod{1000} \equiv \boxed{310} \pmod{1000}.</math> | ||
== See also == | == See also == |
Latest revision as of 16:40, 11 September 2020
Contents
Problem
There are two distinguishable flagpoles, and there are flags, of which are identical blue flags, and are identical green flags. Let be the number of distinguishable arrangements using all of the flags in which each flagpole has at least one flag and no two green flags on either pole are adjacent. Find the remainder when is divided by .
Solution
Solution 1
The well known problem of ordering elements of a string of elements such that none of the elements are next to each other has solutions. (1)
We generalize for blues and greens. Consider a string of elements such that we want to choose the greens such that none of them are next to each other. We would also like to choose a place where we can divide this string into two strings such that the left one represents the first pole, and the right one represents the second pole, in order up the pole according to position on the string.
However, this method does not account for the fact that the first pole could end with a green, and the second pole could start with a green, since the original string assumed that no greens could be consecutive. We solve this problem by introducing an extra blue, such that we choose our divider by choosing one of these blues, and not including that one as a flag on either pole.
From (1), we now have ways to order the string such that no greens are next to each other, and ways to choose the extra blue that will divide the string into the two poles: or orderings in total.
However, we have overcounted the solutions where either pole has no flags, we have to count these separately. This is the same as choosing our extra blue as one of the two ends, and ordering the other blues and greens such that no greens are next to each other: for a total of such orderings.
Thus, we have orderings that satisfy the conditions in the problem: plugging in and , we get .
Solution 2
Split the problem into two cases:
Case 1 - Both poles have blue flags: There are 9 ways to place the 10 blue flags on the poles. In each of these configurations, there are 12 spots that a green flag could go. (either in between two blues or at the tops or bottoms of the poles) Then, since there are 9 green flags, all of which must be used, there are possiblities for each of the 9 ways to place the blue flags. Total: possibilities.
Case 2 - One pole has no blue flags: Since each pole is non empty, the pole without a blue flag must have one green flag. The other pole has 10 blue flags and, by the argument used in case 1, there are possibilities, and since the poles are distinguishable, there are a total of possiblities for this case.
Finally, we have as our answer.
Solution 3 (mad bash)
Call the two flagpoles Flagpole A and Flagpole B. Case 1: Flag distribution: 1|18 B|variable: ways G|variable: ways Case 2: Flag distribution: 2|17 BB|variable: way BG|variable (two ways to arrange the first sequence): ways GG|variable: can't happen Case 3: Flag distribution: 3|16 BBB|variable: can't happen BBG|variable (3 legal ways to arrange): ways GBG|variable (1 legal way to arrange): ways GGG|variable: clearly can't happen Case 4: Flag distribution: 4|15 BBBB|variable: can't happen BBBG|variable (4 legal ways to arrange): ways GBBG|variable (3 legal ways to arrange): ways GGBG|variable: can't happen GGGG|variable: can't happen Case 5: Flag distribution: 5|14 BBBBB|variable: can't happen BBBBG|variable: can't happen BBGBG|variable (6 legal ways to arrange): ways GBGBG|variable (1 legal way to arrange): ways GGGGB|variable: can't happen GGGGG|variable: can't happen Case 6: Flag distribution: 6|13 BBBBBB|variable: can't happen BBBBBG|variable: can't happen BBBBGG|variable (10 legal ways to arrange): ways BBBGGG|variable (4 legal ways to arrange): ways BBGGGG|variable: can't happen BGGGGG|variable: can't happen GGGGGG|variable: can't happen Case 7: Flag distribution: 7|12 BBBBBBB|variable: can't happen BBBBBBG|variable: can't happen BBBBBGG|variable: can't happen BBBBGGG|variable (10 legal ways to arrange): ways BBBGGGG|variable (1 legal way to arrange): ways rest can't happen Case 8: Flag distribution: 8|11 BBBBBBBB|variable: can't happen BBBBBBBG|variable: can't happen BBBBBBGG|variable: can't happen BBBBBGGG|variable (20 legal ways to arrange): ways BBBBGGGG|variable: (5 legal ways to arrange): ways others can't happen Case 9: Flag distribution: 9|10 BBBBBGGGG|variable (15 legal ways to arrange): ways BBBBGGGGG|variable (1 legal way to arrange): ways others can't happen So our total number of ways is times (since the two flagpoles are distinguishable) which is ways. We have to find the last 3 digits, so our final answer is .
Solution by fidgetboss_4000
Note: Do not attempt unless you are good at bashing.
Solution 4 (Vandermonde's)
Let the number of blue flags on the first flag be and define similarly. Now note the bound Now we cannot have any two consecutive greens. This condition is the same as "there are spaces between each blue flag. How many ways can we put the green flags in the spaces?" Therefore given blue flags on the first flag and green flags on the first flag we have a total of Similarly there are ways for the second flag. Summing over all possible possibilites we see the sum is Swapping the sum we have Then applying vandermondes yields so the sum is However, this answer overcounts. We cannot have no flags on a pole, so we subtract since we can have empty flags for the first or second flag. Therefore the answer is
See also
2008 AIME II (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 |
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