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

#Combinatorics

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Harsh Shrivastava @Nihar Mahajan

Adarsh Kumar - 5 years, 4 months ago

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 - 5 years, 4 months ago

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

A Former Brilliant Member - 5 years, 4 months ago

Correct, prove it.

Harsh Shrivastava - 5 years, 4 months ago

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

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 - 5 years, 4 months ago

@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!.

A Former Brilliant Member - 5 years, 4 months ago

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 - 5 years, 4 months ago

@Harsh Shrivastava – "Divide by 2"?

Adarsh Kumar - 5 years, 4 months ago

@Adarsh Kumar – Errr fixed typo.

Harsh Shrivastava - 5 years, 4 months ago

@Harsh Shrivastava – Try the second one I have posted!

Adarsh Kumar - 5 years, 4 months ago

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

A Former Brilliant Member - 5 years, 4 months ago

@A Former Brilliant Member – And I think that is because the groups are identical,so we don't need to permute them.

Adarsh Kumar - 5 years, 4 months ago

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

Adarsh Kumar - 5 years, 4 months ago

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

Adarsh Kumar - 5 years, 4 months ago

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 - 5 years, 4 months ago

No,sorry,just go through the discussion once.

Adarsh Kumar - 5 years, 4 months ago

Is the answer correct?

Harsh Shrivastava - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$