Difference between revisions of "2024 AIME II Problems/Problem 9"
(→Solution 1) |
(→Solution 1) |
||
| Line 7: | Line 7: | ||
The problem says "some", so not all cells must be occupied. | The problem says "some", so not all cells must be occupied. | ||
We start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is <math>1</math> way to select <math>5</math> cells with black chips. Because of the 2nd condition, there can be no white, and the grid must be all black- <math>1</math> way . There are <math>5</math> ways to select 4 cells with black chips. We now consider the row that does not contain a black chip. The first cell must be blank, and the remaining <math>4</math> cells have <math>2^4-1</math> different ways(<math>-1</math> comes from all blank). This gives us <math>75</math> ways. Notice that for 3,2 or 1 black chips on the left there is a pattern. Once the first blank row is chosen, the rest of the blank rows must be ordered similarly. For example, with 2 black chips on the left, there will be 3 blank rows. There are 15 ways for the first row to be chosen, and the following 2 rows must have the same order. Thus, The number of ways for 3,2,and 1 black chips is <math>10*15</math>, <math>10*15</math>, <math>5*15</math>. Adding these up, we have <math>1+75+150+150+75 = 451</math>. Multiplying this by 2, we get <math>\boxed{902}</math>. | We start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is <math>1</math> way to select <math>5</math> cells with black chips. Because of the 2nd condition, there can be no white, and the grid must be all black- <math>1</math> way . There are <math>5</math> ways to select 4 cells with black chips. We now consider the row that does not contain a black chip. The first cell must be blank, and the remaining <math>4</math> cells have <math>2^4-1</math> different ways(<math>-1</math> comes from all blank). This gives us <math>75</math> ways. Notice that for 3,2 or 1 black chips on the left there is a pattern. Once the first blank row is chosen, the rest of the blank rows must be ordered similarly. For example, with 2 black chips on the left, there will be 3 blank rows. There are 15 ways for the first row to be chosen, and the following 2 rows must have the same order. Thus, The number of ways for 3,2,and 1 black chips is <math>10*15</math>, <math>10*15</math>, <math>5*15</math>. Adding these up, we have <math>1+75+150+150+75 = 451</math>. Multiplying this by 2, we get <math>\boxed{902}</math>. | ||
| + | ~westwoodmonster | ||
==See also== | ==See also== | ||
Revision as of 20:54, 8 February 2024
There are 25 indistinguishable white chips and 25 indistinguishable black chips, find the number of ways to place some of these chips such that: Every grid contains at most one chip There can only be one colour of chip in any row or column Any additional chip placed will violate one or more of the previous rules.
Solution 1
The problem says "some", so not all cells must be occupied.
We start by doing casework on the column on the left. There can be 5,4,3,2, or 1 black chip. The same goes for white chips, so we will multiply by 2 at the end. There is 
way to select 
cells with black chips. Because of the 2nd condition, there can be no white, and the grid must be all black- way . There are ways to select 4 cells with black chips. We now consider the row that does not contain a black chip. The first cell must be blank, and the remaining 
cells have 
different ways(
comes from all blank). This gives us 
ways. Notice that for 3,2 or 1 black chips on the left there is a pattern. Once the first blank row is chosen, the rest of the blank rows must be ordered similarly. For example, with 2 black chips on the left, there will be 3 blank rows. There are 15 ways for the first row to be chosen, and the following 2 rows must have the same order. Thus, The number of ways for 3,2,and 1 black chips is 
, , 
. Adding these up, we have 
. Multiplying this by 2, we get 
.
~westwoodmonster
See also
| 2024 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
[[Category:]]
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
