**Prove/Disprove**:

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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.

Log in to reply

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

Log in to reply

Comment deleted Feb 03, 2016

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Correct, prove it.

Log in to reply

@Harsh Shrivastava@Nihar Mahajan

Log in to reply

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

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.

Log in to reply

@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!.

Log in to reply

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.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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)

Log in to reply

No,sorry,just go through the discussion once.

Log in to reply

Is the answer correct?

Log in to reply

Comment deleted Feb 03, 2016

Log in to reply

Log in to reply

Comment deleted Feb 03, 2016

Log in to reply

Log in to reply

Log in to reply