   SEARCH HOME Math Central Quandaries & Queries  Question from Fatima, a parent: Six people call them A,B,C,D,E,F are randomly divided into three groups of two,find the probability of the below event(do not impose unwanted ordering among groups) E andF are in the same group I solved it but I have a doubt that it is wrong . My answer is 576 Please help to solve this problem. We have two responses for you

Well first you should realize that your answer can't be correct as a probability is a number between 0 and 1.

Let's think about E&F being in the same group. That means that the other 4 people are divided into 2 groups. How many ways can you do that? You could pick a person in 4 ways and then a second in 3 ways and that would seem to imply 12 ways but picking A then B doesn't give a different answer than picking B then A so we have to divide by 2, i.e. there are (4x3)/2 = 6 ways in total to have E&F together in a group.

To find the probability that you want you need to divide your answer here, namely 6, by the total number of ways to split the 6 people into 3 groups. I'll get you started on that total number. Let's pick two people for the 1st group. You could pick any one of the 6 first and then any one of the remaining 5 to give (maybe) 30 ways to get the first two - is that right or do we need to divide by 2? Once we've finished doing this calculation correctly, we know from above we can split the remaining 4 people into two groups in 6 ways so the total number of ways to split 6 people into 3 groups is (???) x 6.

Hope this gets you on the right track,

Penny

A probability is always a number between zero and 1, so an answer of 576 cannot be correct.

Since the six people are randomly divided into three groups of two, each such division of the six people into three groups of two is equally likely. To find the probability that such a division involves E and F in the same group of two, we will let m = the number of ways to divide the six people into three groups of two so that E and F are in the same group. We will let n = the total number of ways to divide the six people into three groups of two. Then the answer to the problem will be m/n. Since m and n are both positive and m<=n this will give an answer between 0 and 1 inclusive.

We need two ideas from combinatorics to help out. The first one is that there are a! ways to arrange "a" unique individuals and a! is computed as a*(a-1)*(a-2)*...*1 for a>0 and 0!=1 by definition. For example
6!=6*5*4*3*2*1=720. The second idea is that the quantity "a choose b" is the number of ways to choose b individuals from a group of "a" individuals where a>0, b>=0 and a>=b. This is sometimes abbreviated as aCb. For example, 6C2 is the number of ways to choose 2 individuals from 6 unique individuals. aCb is computed as a!/(b!(a-b)!) thus 6C2=6!/(4!2!)=15.

Let's start by computing n which is the total number of ways to divide six people into three groups of two. To count the number of ways we will divide this problem into a sequence of tasks. The number of ways to
complete the bigger problem will be the product of the numbers of ways to complete each task (with an adjustment at the end).

The tasks will be:
1. Randomly select the first group of two from the six individuals (6C2) ways to do this.
2. Randomly select the second group of two from the remaining four individuals (4C2) ways to do this.
3. Randomly select the third group of two from the remaining two individuals (2C2) ways to do this.

The product of the numbers of ways to do tasks 1,2,3 is (6C2)*(4C2)*(2C2).

However, notice that the tasks create a "first group", "second group" and "third group". We don't really care about the arrangement of the groups, for example "AB", "CD", "EF" is really the same as the "EF", "AB", "CD". So using these tasks we have counted each division of the six individuals into three groups of two more than once. Since there are 3! ways to arrange three pairs, then using (6C2)(4C2)(2C2) counts each division of the six individuals into three groups of two 3! times. So the total number of ways to divide six people into three groups of two is n=(6C2)*(4C2)*(2C2)/3! = 15

We can use similar reasoning to count the number of ways to divide the six people into three groups of two so that E and F are in the same group.
1. Realize that E and F are in a group of two (1 way to do this).
2. Randomly select two people from the remaining four for the secondgroup (4C2 ways).
3. Randomly select two people from the remaining two for the third group(2C2 ways).
As above m= 1*4C2*2C2/2! = 3 (EF is always in the first group)

Thus the answer to this probability question is 3/15=1/5=0.2

A shorter way to solve this problem is the following:

Let x be the number of ways to divide four people into two groupings of two people each.

Then n = 6C2*x and m=1*x so the answer is 1*x/6C2*x = 1/5=0.2

Hope this is correct I'm a bit rusty!

Thx,
Lorraine     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.