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

Note by Adarsh Kumar
1 year, 11 months ago

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A follow up question :

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.

Harsh Shrivastava - 1 year, 11 months ago

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

Kalash Verma - 1 year, 11 months ago

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Comment deleted Feb 03, 2016

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@Adarsh Kumar What about the edited answer? @Adarsh Kumar

Kalash Verma - 1 year, 11 months ago

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

Adarsh Kumar - 1 year, 11 months ago

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@Adarsh Kumar Yeah!

Kalash Verma - 1 year, 11 months ago

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Correct, prove it.

Harsh Shrivastava - 1 year, 11 months ago

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

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.

Harsh Shrivastava - 1 year, 11 months ago

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@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!.

Kalash Verma - 1 year, 11 months ago

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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.

Harsh Shrivastava - 1 year, 11 months ago

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

Adarsh Kumar - 1 year, 11 months ago

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

Kalash Verma - 1 year, 11 months ago

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

Adarsh Kumar - 1 year, 11 months ago

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@Harsh Shrivastava "Divide by 2"?

Adarsh Kumar - 1 year, 11 months ago

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@Adarsh Kumar Errr fixed typo.

Harsh Shrivastava - 1 year, 11 months ago

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@Harsh Shrivastava Try the second one I have posted!

Adarsh Kumar - 1 year, 11 months ago

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

Adarsh Kumar - 1 year, 11 months ago

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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)

Nihar Mahajan - 1 year, 11 months ago

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No,sorry,just go through the discussion once.

Adarsh Kumar - 1 year, 11 months ago

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Is the answer correct?

Harsh Shrivastava - 1 year, 11 months ago

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Comment deleted Feb 03, 2016

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@Adarsh Kumar Where is Kalash's solution?

Nihar Mahajan - 1 year, 11 months ago

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Comment deleted Feb 03, 2016

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

Nihar Mahajan - 1 year, 11 months ago

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

Adarsh Kumar - 1 year, 11 months ago

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