# Thread: tom and jerry

1. ## tom and jerry

There is a group of 10 people. We will seperate this group into 5-person groups but we don't want Tom AND Jerry to be in the same group. How many ways are there?

My work
without limitation:
C(9,1)C(7,1)C(5,1)C(4,1)

If we exclude Tom AND Jerry,there are 8 persons.

C(8,4)C(2,1) is my way but book says 240

2. ## Re: tom and jerry

Are the group's numbered? Because if not, there are $\dbinom{10}{5}$ ways to distribute 10 people into two distinguishable groups of five. Suppose a group has both Tom and Jerry in it. There are $\dbinom{8}{3}$ ways to fill the group with 3 more people. But Tom and Jerry could be together in either group. So, there are 2x as many ways that could happen. I get $252-112=140$, which is the same answer as you.

3. ## Re: tom and jerry

Why did you multiplied by 2?

Because there are 2 groups?

4. ## Re: tom and jerry

Correct. If the groups are distinguishable, the two could be together in the first group or the second.

5. ## Re: tom and jerry

Note: If the groups are indistinguishable, then this changes the problem significantly. There are $\dfrac{1}{2}\dbinom{10}{5} = 126$ ways to distribute 10 people into two indistinguishable groups of 5. Then, to ensure that Tom and Jerry are not in the same group, there are $\dbinom{8}{3}$ ways to choose three more people to be in a group with Tom and Jerry (and the other group has the remaining 5 people). Since we did not say which group Tom and Jerry are in, we do not need to divide by 2.

So, there are $126-56 = 70 = \dfrac{1}{2}\dbinom{8}{4}\dbinom{2}{1}$ ways to distribute 10 people into two indistinguishable groups of 5 with Tom and Jerry in separate groups.

Many Thanks

7. ## Re: tom and jerry

Originally Posted by kastamonu
There is a group of 10 people. We will seperate this group into 5-person groups but we don't want Tom AND Jerry to be in the same group. How many ways are there?
This maybe beating a dead horse?
The groupings are distinct: one contains Tom & the other Jerry.
The answer is $\dbinom{8}{4}=70$ HERE
Choose four other than Jerry to go with Tom, the remainder form Jerry's group.

Many Thanks.

9. ## Re: tom and jerry

Originally Posted by kastamonu
There is a group of 10 people. We will seperate this group into 5-person groups but we don't want Tom AND Jerry to be in the same group. How many ways are there?
P.S.
This type of question fall into the category known as unordered partitions.

Say that $N$ is a multiple of $k$ i.e. $N=j\cdot k$. So a collection of $N$ distinct objects can be grouped into $j$ groups of $k$ each.
This is called because the groups are indistinguishable. This can be done in $\dfrac{N!}{(k!)^j(j!)}$ ways.
I have found that this is a difficult concepts to get across to even graduate students.

For example: How many ways can a class of twenty be arranged into four groups of five each?
If that formula is correct then $\dfrac{20!}{(5!)^4(4!}=488864376$ See HERE.

I once heard this another approach given is a talk. I did not believe it at first.
Select anyone in the class. Choose four more from the nineteen left. The select anyone not chosen then choose four more to go with her/him. Repeat one more time and then we will have our four groups.
$\dbinom{19}{4}\cdot \dbinom{14}{4}\cdot\dbinom{9}{4}\cdot\dbinom{4}{4}$
If like me you do not see it look HERE

10. ## Re: tom and jerry

I don't know any of these formulas. Maybe they show them in Engineering Faculties. But there may be a logical way of solving combination problems without formulas.

11. ## Re: tom and jerry

Let's suppose that we don't have any experience. We don't know any formula. And this problem is before us. How should you have solved that?

12. ## Re: tom and jerry

Originally Posted by kastamonu
Let's suppose that we don't have any experience. We don't know any formula. And this problem is before us. How should you have solved that?
Plato said it is difficult to understand where that formula comes from. We would not be able to explain it in this forum. That is something that would require classroom instruction.