# Counting problem

## Homework Statement

100 hundred people are to be divided into 10 discussion groups with 10 people in each group
how many ways can this be done.

## The Attempt at a Solution

So if we think of it as people on a 10 by 10 grid their are 100! ways of populating the grid and then 10! ways or rearranging the columns and 10! ways of rearranging the rows.
so would the answer be
$\frac{100!}{10!^{10}10!}$
I have 10!^10 because there are ten columns and each of those 10 columns can be arranged 10! ways. And the other 10! on the bottom because I could rearrange those rows 10! ways. And I divide 100! by them because we are over counting. If we switch the people around in the group it is still a distinct group.

## Answers and Replies

lanedance
Homework Helper
didn't totally follow the re-arranging of the cols, so as another way for the first group we choose 10 from 100 without order
$$\frac{100!}{10!90!}$$

for the 2nd group we choose 10 from 90 without order
$$\frac{90!}{10!80!}$$

so putting these together for both groups we get
$$\frac{100!}{10!90!} \frac{90!}{10!80!} = \frac{100!}{10!^280!}$$

and so on, once finished counting the groups we need to account for the different ways to arrange ten groups which represents the repeated counts -

but its starting to look pretty similar to your answer...

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lanedance
Homework Helper
also another check would be to see if you formula generalise to n^2 people in n groups and check for n = 2 and n= 3

lanedance
Homework Helper
ok - so i get your idea now, but think it needs a little tweek.. the rows got me

100! ways to populate a grid, where the columns represent a group
- 10! ways to rearrange within a single column and 10 columns gives a total (10!)^10 ways to rearrange within columns
- there is also 10! different ways to arrange the columns themsleves
So in all there is (10!)^11 ways to rearrange the grid for a given combination that leads to the same group structure

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thanks for your responses, ya i think my answer makes sense.