(New page: The number of ways to distribute 5 indistinguishable objects into 3 indistinguishable boxes is essentially the number of ways to find solutions to <math>x_1+x_2+x_3 = 5</math>, where <math...) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 1: | Line 1: | ||
The number of ways to distribute 5 indistinguishable objects into 3 indistinguishable boxes is essentially the number of ways to find solutions to <math>x_1+x_2+x_3 = 5</math>, where <math>x_1</math>, <math>x_2</math>, and <math>x_3</math> are non-negative integers. Since the boxes are indistinguishable, solution (a,b,c) is no different than (b,a,c), (c,a,b), etc. Therefore, our question is really how many ways we can split five into three parts, order doesn't matter. | The number of ways to distribute 5 indistinguishable objects into 3 indistinguishable boxes is essentially the number of ways to find solutions to <math>x_1+x_2+x_3 = 5</math>, where <math>x_1</math>, <math>x_2</math>, and <math>x_3</math> are non-negative integers. Since the boxes are indistinguishable, solution (a,b,c) is no different than (b,a,c), (c,a,b), etc. Therefore, our question is really how many ways we can split five into three parts, order doesn't matter. | ||
The possible distributions are (0,0,5), (0,2,3), (0,1,4), (1,2,2), and (1,1,3). Thus, there are five possible ways. | The possible distributions are (0,0,5), (0,2,3), (0,1,4), (1,2,2), and (1,1,3). Thus, there are five possible ways. | ||
| + | |||
| + | --[[User:Zhao14|Zhao14]] 19:09, 28 September 2008 (UTC) | ||
| + | zhao14@purdue.edu | ||
Latest revision as of 15:10, 28 September 2008
The number of ways to distribute 5 indistinguishable objects into 3 indistinguishable boxes is essentially the number of ways to find solutions to $ x_1+x_2+x_3 = 5 $, where $ x_1 $, $ x_2 $, and $ x_3 $ are non-negative integers. Since the boxes are indistinguishable, solution (a,b,c) is no different than (b,a,c), (c,a,b), etc. Therefore, our question is really how many ways we can split five into three parts, order doesn't matter. The possible distributions are (0,0,5), (0,2,3), (0,1,4), (1,2,2), and (1,1,3). Thus, there are five possible ways.
--Zhao14 19:09, 28 September 2008 (UTC) zhao14@purdue.edu
