**Prove/Disprove**:

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

**Prove/Disprove**:

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

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TopNewestA 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, 3 months ago

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– Kalash Verma · 1 year, 3 months ago

(a+b+c)!/3!a!b!c!Log in to reply

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– Kalash Verma · 1 year, 3 months ago

What about the edited answer? @Adarsh KumarLog in to reply

– Adarsh Kumar · 1 year, 3 months ago

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

– Kalash Verma · 1 year, 3 months ago

Yeah!Log in to reply

– Harsh Shrivastava · 1 year, 3 months ago

Correct, prove it.Log in to reply

@Harsh Shrivastava@Nihar Mahajan – Adarsh Kumar · 1 year, 3 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, 3 months ago

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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, 3 months ago

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I forgot that we also have to divide by 3! since there are repetitions. – Harsh Shrivastava · 1 year, 3 months ago

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– Adarsh Kumar · 1 year, 3 months ago

Could you elaborate a bit more?What repetitions are you talking about?Log in to reply

– Kalash Verma · 1 year, 3 months ago

It will be divided by 3! .There was a typo in my commentLog in to reply

– Adarsh Kumar · 1 year, 3 months ago

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

– Adarsh Kumar · 1 year, 3 months ago

"Divide by 2"?Log in to reply

– Harsh Shrivastava · 1 year, 3 months ago

Errr fixed typo.Log in to reply

– Adarsh Kumar · 1 year, 3 months ago

Try the second one I have posted!Log in to reply

– Adarsh Kumar · 1 year, 3 months ago

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

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, 3 months ago

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– Adarsh Kumar · 1 year, 3 months ago

No,sorry,just go through the discussion once.Log in to reply

– Harsh Shrivastava · 1 year, 3 months ago

Is the answer correct?Log in to reply

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– Nihar Mahajan · 1 year, 3 months ago

Where is Kalash's solution?Log in to reply

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– Nihar Mahajan · 1 year, 3 months ago

I haven't given enough time for this problem . I will revisit this problem later.Log in to reply

– Adarsh Kumar · 1 year, 3 months ago

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