Ordered and Unordered partitions

mathdaemon

New member
Joined
Dec 26, 2012
Messages
35
Hello everyone,

I am a bit confused as to how to solve problems based on ordered and unordered partitions in Cominatorics.

Ok, so I have two questions

1) How many ways can 10 persons be divided into 3 teams such that Team-1 contains 3, Team-2 contains 2 and Team-3 contains 5 members respectively?

The answer to this is 10!/(3! X 2! X 5!). This can be obtained from the permutations with repeatition formula n!/(a! X b! X c!...).
I don't understand how this formula can be used here because persons are distinct.

2) How many ways can 10 persons be divided into 5 teams of 2 each?

My incorrect solution is .... 1st team chosen in 10 X 9 ways, 2nd in 8 X 7 ways and so on.
Why is the solution incorrect.
What should be the logic of the correct solution(10! / (5 X 2! X 5!))

3) In the solution above 5! is for arranging teams among themselves in five ways.
Is the same logic valid if in question 1) the sizes of 2(or more) teams are the same?

I hope my questions are understandable.
Please do help.
Thanks in advance
 
Last edited:
1) How many ways can 10 persons be divided into 3 teams such that Team-1 contains 3, Team-2 contains 2 and Team-3 contains 5 members respectively?

The answer to this is 10!/(3! X 2! X 5!). This can be obtained from the permutations with repeatition formula n!/(a! X b! X c!...).
I don't understand how this formula can be used here because persons are distinct.

2) How many ways can 10 persons be divided into 5 teams of 2 each?

My incorrect solution is .... 1st team chosen in 10 X 9 ways, 2nd in 8 X 7 ways and so on.
Why is the solution incorrect.
What should be the logic of the correct solution(10! / (5 X 2! X 5!))

3) In the solution above 5! is for arranging teams among themselves in five ways.
Is the same logic valid if in 1) the sizes of 2(or more) teams the same?
Sorry but this is a hurried response.
The string \(\displaystyle "AaBbCcDd"\) can be rearranged in \(\displaystyle 8!\) ways because we distinguish upper from lower case letters. However, if we totally disregard cases then the string can be rearranged in \(\displaystyle \dfrac{8!}{(2!)^4(4!)}\) ways. That is the number of ways that eight people can be on four teams of two each.
 
Last edited:
Thanks for replying.

In how many ways 10 persons can be divided into 5 teams of 2 each
10!/(2!)^5=113400 but the answer given is 945.

Also, can you please throw some light on the logic of solving ordered partition problems(question 1 and 3).

Thanks
 
Thanks for replying.

In how many ways 10 persons can be divided into 5 teams of 2 each
10!/(2!)^5=113400 but the answer given is 945.

Also, can you please throw some light on the logic of solving ordered partition problems(question 1 and 3).

Thanks
See my edit above: also divide by 5!
 
First, in situations such as this you have to decide whether order matters or order doesn't matter. In this case, a team consisting of persons (a, b, c) would be the same team as one consisting of (b, c, a). So order doesn't matter for this case. In math parlance what we want is a combination rather than a permutation where the order does matter.

Another decision which is needed is whether replacement is allowed. In this case, you obviously can't chose person a for two different teams so replacement isn't allowed.

A nice, IMO, write up on this is given at
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

So, let's tackle the first problem
1) How many ways can 10 persons be divided into 3 teams such that Team-1 contains 3, Team-2 contains 2 and Team-3 contains 5 members respectively?

For Team 1 we want to choose three people out of ten where order doesn't matter. The way I learned this was to use the notation nCr (or nCr) and read it as n Choose r where
nCr = \(\displaystyle \dfrac{n!}{r! (n-r)!}\)

For our particular case we have 10 Choose 3 which evaluates to
10C3 = \(\displaystyle \dfrac{10!}{(7!) (3!)}\) = \(\displaystyle \dfrac{(10) (9) (8)} {(1) (2) (3)}\) = 120

For Team 2, there are only 7 persons left and we need to choose 2 of them, so we have
7C2 = 21

For Team 3, there are only 5 persons left and we have to choose all 5, so we have
5C5 = 1

Now how many combinations can we make of these? We start with any of 120, then any of 21 and finally just the 1 is left, 120*21*1 = 2520.

Another way to look at this is that we are going to choose 10 people without replacement. That gives the \(\displaystyle 10!\). But we have to reduce that number by the number of ways we can arrange the three groups because order doesn't matter. Well, we can arrange the three group \(\displaystyle 3!\) ways, the 2 group \(\displaystyle 2!\) ways and the 5 group \(\displaystyle 5!\) ways. So our final answer is
\(\displaystyle \dfrac{10!}{(2!) (3!) (5!)}\) = 2520.
This also shows that the order in which we choose our teams of 3, 2, and 5 doesn't matter. We could have chosen the 5 person teams first, then the 2 person teams, and finally the three person team.
 
Last edited:
Top