# certain number of people arranged in several groups

• July 9th 2014, 07:27 AM
delso
certain number of people arranged in several groups
Find the number of ways of 9 people can be divided into two groups of 6 people and 3 people.

my working is there are 6! arrangement for 6 people and 3! arrangment for 3 people. then the group of 3 people can be placed at right or left only . so that the 6 people form another group will not be seperated. so my working = 3! x 6! x2 = 8640..

the ans given is only 84
• July 9th 2014, 08:07 AM
SlipEternal
Re: certain number of people arranged in several groups
Let's use smaller numbers. Let's say there are only 3 people, and we want to count the number of ways to put them in a group of 2 and a group of 1. According to your method, this is $2!\cdot 1!\cdot 2 = 4$

Suppose the people are A, B, and C. Let's list off each possible grouping:
{A,B} and {C}
{A,C} and {B}
{B,C} and {A}

We could also permute the two people in each group, but we get the same group. The order doesn't matter. So these are the only three. See if you can figure out why your counting method overcounted.

Back to the original problem: let's look at permutations of the nine people. The first six people (in the permutation) go to one group and the last three go to another. For a given permutation, let's say [A,B,C,D,E,F,G,H,I], we obtain the groups {A,B,C,D,E,F} and {G,H,I}. However, any permutation where the letters A through F appear in the first six positions and G through I appear in the last three gives this same grouping. So, among those permutations, we get this same grouping how many times? It is the number of ways to permute 6 elements times the number of ways to permute 3 elements. Since this is true for EVERY pair of groups, we are counting the groups $6!3!$ times each. Hence, the number of groups is $\dfrac{9!}{6!3!} = \binom{9}{6} = \binom{9}{3} = {{}_9C_6} = {{}_9C_3}= \dfrac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1} = 3\cdot 4\cdot 7 = 84$
• July 9th 2014, 09:32 PM
delso
Re: certain number of people arranged in several groups
here's another example for my textbook.

6 men and 4 women are to be seated in a row. Find the njumer of ways of arrange them if all the women sit together in the group.

the ans given is 6! x 4! x 2 =34560

the reasoning is same as i did it the #1 post
• July 10th 2014, 04:58 AM
HallsofIvy
Re: certain number of people arranged in several groups
Quote:

Originally Posted by delso
Find the number of ways of 9 people can be divided into two groups of 6 people and 3 people.

my working is there are 6! arrangement for 6 people and 3! arrangment for 3 people. then the group of 3 people can be placed at right or left only . so that the 6 people form another group will not be seperated. so my working = 3! x 6! x2 = 8640..

You seem to have completely misunderstood the questions. 3! is the number of different orders for 3 given people. But "order" is not relevant here.

Quote:

the ans given is only 84
• July 10th 2014, 07:34 AM
delso
Re: certain number of people arranged in several groups
Quote:

Originally Posted by HallsofIvy
You seem to have completely misunderstood the questions. 3! is the number of different orders for 3 given people. But "order" is not relevant here.

why the arrangement of people in the group is not important for this question? it never stated so i am always confused when i encounter this type of question.
• July 10th 2014, 07:44 AM
delso
Re: certain number of people arranged in several groups
Quote:

Originally Posted by HallsofIvy
You seem to have completely misunderstood the questions. 3! is the number of different orders for 3 given people. But "order" is not relevant here.

why the arrangement of people in the group is not important for this question? it never stated so i am always confused when i encounter this type of question.

by the way, can I do it in the way that 9c3 x 6c6=84? but by doing so means that i choose 3 person first , then only the 6 person?
• July 10th 2014, 08:34 AM
SlipEternal
Re: certain number of people arranged in several groups
You are not choosing first or second. The choice is made at the same time. If you choose three people from nine, six people remain. If you choose six people from nine, three people remain. No matter how you pick the six people for the group of six, there will only be three people left for the group of three. No matter how you choose the three people for the group of three, there will be six people left for the group of six.
• July 10th 2014, 09:32 AM
HallsofIvy
Re: certain number of people arranged in several groups
Quote:

Originally Posted by delso
why the arrangement of people in the group is not important for this question?

Because the 3!= 6 orders
{Anne, Bill, Charlie}
{Anne, Charlie, Bill}
{Bill, Anne, Charlie}
{Bill, Charlie, Anne}
{Charlie, Anne, Bill}
{Charlie, Bill, Anne}
are all exactly the same 3 person committee. You do NOT want to count them as different.

If the problem had said "chosen as President, Vice-President, Treasurer", then those would be different: the first might be interpreted as "Anne is president, Bill is vice-president, Charlie is treasurer", the second as "Anne is president, Charlie is vice-president, Bill is Treasurer", etc. In that case each of those six "permutations " of the same three people is a different answer.

Quote:

it never stated so i am always confused when i encounter this type of question.

by the way, can I do it in the way that 9c3 x 6c6=84? but by doing so means that i choose 3 person first , then only the 6 person?
• July 11th 2014, 01:30 AM
delso
Re: certain number of people arranged in several groups
thanks , everything clear now!