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# 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 $$\dfrac{(3p)!}{(p!)^3}$$.

1 year, 1 month 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. · 1 year, 1 month ago

(a+b+c)!/3!a!b!c! · 1 year, 1 month ago

Comment deleted Feb 03, 2016

Yup!That is the correct one!I hope you know why and how you get this! · 1 year, 1 month ago

Yeah! · 1 year, 1 month ago

Correct, prove it. · 1 year, 1 month 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. · 1 year, 1 month 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!. · 1 year, 1 month 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. · 1 year, 1 month ago

Could you elaborate a bit more?What repetitions are you talking about? · 1 year, 1 month ago

It will be divided by 3! .There was a typo in my comment · 1 year, 1 month ago

And I think that is because the groups are identical,so we don't need to permute them. · 1 year, 1 month ago

"Divide by 2"? · 1 year, 1 month ago

Errr fixed typo. · 1 year, 1 month ago

Try the second one I have posted! · 1 year, 1 month ago

I am sorry to say but that is incorrect,just think it up for some time.It is a very common mistake. · 1 year, 1 month 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) · 1 year, 1 month ago

No,sorry,just go through the discussion once. · 1 year, 1 month ago

Is the answer correct? · 1 year, 1 month ago

Comment deleted Feb 03, 2016

Where is Kalash's solution? · 1 year, 1 month ago

Comment deleted Feb 03, 2016

I haven't given enough time for this problem . I will revisit this problem later. · 1 year, 1 month ago