Suppose a number, say \(n\) is given.

How to find number of solutions of the equation?

\(x \times y \times z = n\)

Please help.

And also what is the best way to approach combinatorics problems in JEE Advanced?

What are the most important topics in Combinatorics that I have to cover?

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:

TopNewestThe way i learnt was, suppose \(xyz=2^{3}\times 3^{2}\)

using the method, Number of ways to distribute \(r\) identical objects among \(n\) different objects is \(\huge{^{r+n-1}C_{n-1}}\)

Thus we want to distribute three identical 2's and two identical three's among three objects.

Required ways \(^{3+3-1}C_{3-1}\times ^{2+3-1}C_{3-1}\)

hope it helps

Log in to reply

Lovely solution!I had also thought of one,but it was a bit more complex!

Log in to reply

My friend asked me this question:

No of solution for the equation:

\(x \times y \times z \times w= 210 \)

I struggled to get the answer. Could you try this?

Log in to reply

I am getting 256, u see we have to distribute one 2 among 4 objects, and the method above i posted gives \(^{4}C_{3}=4\). Hence the answer is \(4^{4}=256\)

Log in to reply

Log in to reply

this

can u help me withLog in to reply

Log in to reply

Log in to reply

Hence,

\(\lambda_{K_{\alpha}} - \frac{hc}{e\frac{V}{2}} = 4 \times (\lambda_{K_{\alpha}} - \frac{hc}{eV} )\)

Log in to reply

Then use Moseley's equation.

Log in to reply

Log in to reply

Log in to reply

210 can be written as 2 x 3 x 5 x 7.Hence if none of the given variables are 1,then there exist 24 solutions(4!).Then if one of the variables is 1,(4 ways)then we are left with 2,3,5,7.We have to divide these numbers in three parts,which can be done in C(3,2) ways : 3 gaps 2 bars.So total number of ways with one variable as 1 is 4 x 3=12. If two variables are 1 then we have to divide 2,3,5,7 between two integers which can be done in C(3,1) ways so total = 6(4 c 2) x 3=18 ways.Then last case has 4 solutions.Total=4+18+24+12=58 ways.

Log in to reply

Log in to reply

Log in to reply

Positive Integer solutions.Log in to reply

Log in to reply

Log in to reply