# Thread: Forming two groups of 4 students from a class of 3 boys and 4 girls, with retsriction

1. ## Forming two groups of 4 students from a class of 3 boys and 4 girls, with retsriction

A teacher is to form two groups of 4 students from a class of 3 boys and 4 girls for consultation session. three of the girls, ivy, tamie and cassy are good frens. find the prob that either ivy or tamie is in the same grp as cassy.

i remove cassy from the group 1st. then (2C1x 5C2)/7C3.

but this is wrong. can u guys help me out?

2. ## Re: probability

Hello, helloying!

A teacher is to form two groups of 3 students from a class of 3 boys and 4 girls.
Three of the girls, Ivy, Tamie and Cassy are good friends.
Find the probability that either Ivy or Tamie is in the same group as Cassy.

You are ignoring most of the given information.
And you are misreading the question.

There are 7 students.
We select 6 of them: .$\displaystyle _7C_3 \:=\:{7\choose3} \:=\:35$ ways.
Then we divide them into two groups of 3 students: .$\displaystyle {6\choose3,3} \:=\:20$ ways.
. . Hence, there are: .$\displaystyle 35\cdot20 \:=\:700$ possible groupings.

How many groupings have: .$\displaystyle \begin{Bmatrix}\text{(Cassy, Ivy)} \\ \text{(Cassy, Tamie)} \\ \text{(Cassy, Ivy, Tamie)} \end{Bmatrix}$ in the same group?

3. ## Re: probability

Originally Posted by helloying
A teacher is to form two groups of 4 students from a class of 3 boys and 4 girls for consultation session. three of the girls, ivy, tamie and cassy are good frens. find the prob that either ivy or tamie is in the same grp as cassy. (2C1x 5C2)/7C3.
I have a somewhat different understanding of this question that does Soroban. I read it as your solution suggests: two groups of two students each.

There are $\displaystyle \binom{7}{4}\frac{4!}{2(2!)^2}=105$ ways to pick two groups of two students each.

There are $\displaystyle 2\binom{5}{2}=20$ ways for Ivy and Cassy or Tamie and Cassy to be in the same one of those 105 groupings.

4. ## Re: probability

Opps sorry i realise my qn is wrong. there should be 5 girls. so there should be two groups of 4 students each. the ans given is 5/14.

5. ## Re: probability

Originally Posted by helloying
Opps sorry i realise my qn is wrong. there should be 5 girls. so there should be two groups of 4 students each. the ans given is 5/14.
I am sorry to tell that I failed to see this correction.
Also sorry to tell you that the proposed answer is incorrect.
There has been a pattern in your posts, the given answer is often wrong.
In this case the correct answer is $\displaystyle \frac{5}{7}.$

Here are two ways to see that.
There are $\displaystyle \frac{8!}{(4!)^2(2)}=35$ ways to form two groups of four each from this collection.

There are $\displaystyle 2\binom{6}{2}-\binom{5}{1}$ ways for Cassy & Tamie or Cassy & Ivy to be in the same group.

Here is the second way. There are $\displaystyle \binom{5}{3}$ ways that Cassy, Ivy, & Tamie are not in the same group.

Now you do the calculations. In either case the answer is $\displaystyle \frac{5}{7}.$