Prove/Disprove Combinatorics 1

Prove/Disprove:

If 3p3p distinct elements are divided in three groups each containing pp elements.then the number of divisions is (3p)!(p!)3\dfrac{(3p)!}{(p!)^3}.

Note by Adarsh Kumar
3 years, 6 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 - 3 years, 6 months ago

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

Kalash Verma - 3 years, 6 months ago

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Correct, prove it.

Harsh Shrivastava - 3 years, 6 months ago

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(3pp)(2pp)(pp)=(3p)!(p!)3\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 - 3 years, 6 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 - 3 years, 6 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 - 3 years, 6 months ago

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@Harsh Shrivastava "Divide by 2"?

Adarsh Kumar - 3 years, 6 months ago

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@Adarsh Kumar Errr fixed typo.

Harsh Shrivastava - 3 years, 6 months ago

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@Harsh Shrivastava Try the second one I have posted!

Adarsh Kumar - 3 years, 6 months ago

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

Kalash Verma - 3 years, 6 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 - 3 years, 6 months ago

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

Adarsh Kumar - 3 years, 6 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 - 3 years, 6 months ago

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Without any restriction , the 3p3p elements can be arranged in (3p)!(3p)! . Now we need to compute the total repetition factor. In each group , the pp elements can be permuted in p!p! ways , so by multiplication principle in three groups, total repetition factor =(p!)3=(p!)^3 . Hence , number of divisions =(3p)!(p!)3=\frac{(3p)!}{(p!)^3} .

(I may be wrong)

Nihar Mahajan - 3 years, 6 months ago

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No,sorry,just go through the discussion once.

Adarsh Kumar - 3 years, 6 months ago

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Is the answer correct?

Harsh Shrivastava - 3 years, 6 months ago

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