# Prove/Disprove Combinatorics 1

Prove/Disprove:

If $$3p$$ distinct elements are divided in three groups each containing $$p$$ elements.then the number of divisions is $$\dfrac{(3p)!}{(p!)^3}$$.

2 years, 5 months ago

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Let there be n = a+b+c distinct things.Find the number of ways of dividing these n things in groups containing a,b,c things.

- 2 years, 5 months ago

(a+b+c)!/3!a!b!c!

- 2 years, 5 months ago

Comment deleted Feb 03, 2016

- 2 years, 5 months ago

Yup!That is the correct one!I hope you know why and how you get this!

- 2 years, 5 months ago

Yeah!

- 2 years, 5 months ago

Correct, prove it.

- 2 years, 5 months ago

- 2 years, 5 months ago

$$\dbinom{3p}{p} \dbinom{2p}{p} \dbinom{p}{p} = \dfrac{(3p)!}{(p!)^{3}}$$

Out of 3p things, we can first choose p things, then outta remaining 2p things, we can choose p things and out of remaining p things we can choose p things in 1 one way only.

- 2 years, 5 months ago

@Nihar Mahajan and@Harsh Shrivastava try to find the mistake using the example.

Let p =1(elements be 1,2,3) then n.o of different groups is 1 but according to your generalization its coming 3!.

- 2 years, 5 months ago

Yup , I'll rectify my solution later, a bit buzy now,

I forgot that we also have to divide by 3! since there are repetitions.

- 2 years, 5 months ago

Could you elaborate a bit more?What repetitions are you talking about?

- 2 years, 5 months ago

It will be divided by 3! .There was a typo in my comment

- 2 years, 5 months ago

And I think that is because the groups are identical,so we don't need to permute them.

- 2 years, 5 months ago

"Divide by 2"?

- 2 years, 5 months ago

Errr fixed typo.

- 2 years, 5 months ago

Try the second one I have posted!

- 2 years, 5 months ago

I am sorry to say but that is incorrect,just think it up for some time.It is a very common mistake.

- 2 years, 5 months ago

Without any restriction , the $$3p$$ elements can be arranged in $$(3p)!$$ . Now we need to compute the total repetition factor. In each group , the $$p$$ elements can be permuted in $$p!$$ ways , so by multiplication principle in three groups, total repetition factor $$=(p!)^3$$ . Hence , number of divisions $$=\frac{(3p)!}{(p!)^3}$$ .

(I may be wrong)

- 2 years, 5 months ago

No,sorry,just go through the discussion once.

- 2 years, 5 months ago

- 2 years, 5 months ago

Comment deleted Feb 03, 2016

Where is Kalash's solution?

- 2 years, 5 months ago

Comment deleted Feb 03, 2016

I haven't given enough time for this problem . I will revisit this problem later.

- 2 years, 5 months ago

Ohk,sure!Let us delete some of our not-so-important comments.

- 2 years, 5 months ago