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.

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 $=\frac{(3p)!}{(p!)^3}$ .

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## Comments

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TopNewest@Harsh Shrivastava@Nihar Mahajan

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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.

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

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Correct, prove it.

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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.

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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!.

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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.

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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.

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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)

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No,sorry,just go through the discussion once.

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Is the answer correct?

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