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?

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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 – Tanishq Varshney · 2 years ago

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– Adarsh Kumar · 2 years ago

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

No of solution for the equation:

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

I struggled to get the answer. Could you try this? – Vishwak Srinivasan · 2 years ago

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– Tanishq Varshney · 2 years ago

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

– Vishwak Srinivasan · 2 years ago

Aha. Okay. Thanks bro!Log in to reply

this – Tanishq Varshney · 2 years ago

can u help me withLog in to reply

– Vishwak Srinivasan · 2 years ago

Dude, in that question, \(K_{\alpha} \) is a characteristic ray and does not depend on the Potential Difference applied.Log in to reply

– Tanishq Varshney · 2 years ago

can u explain in detail, plzLog in to reply

Hence,

\(\lambda_{K_{\alpha}} - \frac{hc}{e\frac{V}{2}} = 4 \times (\lambda_{K_{\alpha}} - \frac{hc}{eV} )\) – Vishwak Srinivasan · 2 years ago

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Then use Moseley's equation. – Vishwak Srinivasan · 2 years ago

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– Tanishq Varshney · 2 years ago

thanks for the helpLog in to reply

– Vishwak Srinivasan · 2 years ago

No problem dude. Hope you got the answer.Log in to reply

– Adarsh Kumar · 2 years ago

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

– Vishwak Srinivasan · 2 years ago

Adarsh, wrong answer.Log in to reply

– Adarsh Kumar · 2 years ago

Do you know the answer?Do we have to find integer solutions?Log in to reply

Positive Integer solutions. – Vishwak Srinivasan · 2 years agoLog in to reply

– Adarsh Kumar · 2 years ago

Even i got 220 on the second try!Log in to reply

– Adarsh Kumar · 2 years ago

oh!ok let me see where i have made a mistake!Log in to reply