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Please I need help....Combinatorics

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?

Note by Vishwak Srinivasan
2 years ago

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The 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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@Tanishq Varshney Lovely solution!I had also thought of one,but it was a bit more complex! Adarsh Kumar · 2 years ago

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@Tanishq Varshney 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? Vishwak Srinivasan · 2 years ago

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@Vishwak Srinivasan 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 \(4^{4}=256\) Tanishq Varshney · 2 years ago

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@Tanishq Varshney Aha. Okay. Thanks bro! Vishwak Srinivasan · 2 years ago

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@Vishwak Srinivasan can u help me with this Tanishq Varshney · 2 years ago

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@Tanishq Varshney Dude, in that question, \(K_{\alpha} \) is a characteristic ray and does not depend on the Potential Difference applied. Vishwak Srinivasan · 2 years ago

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@Vishwak Srinivasan can u explain in detail, plz Tanishq Varshney · 2 years ago

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@Tanishq Varshney \(\lambda_{K_{\alpha}} \)does not change with change in potential difference.

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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@Vishwak Srinivasan Now solve for \(\lambda_{K_{\alpha}} \).

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

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@Vishwak Srinivasan thanks for the help Tanishq Varshney · 2 years ago

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@Tanishq Varshney No problem dude. Hope you got the answer. Vishwak Srinivasan · 2 years ago

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@Vishwak Srinivasan 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. Adarsh Kumar · 2 years ago

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@Adarsh Kumar Adarsh, wrong answer. Vishwak Srinivasan · 2 years ago

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@Vishwak Srinivasan Do you know the answer?Do we have to find integer solutions? Adarsh Kumar · 2 years ago

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@Adarsh Kumar The Answer is 256. I got 220. I've been breaking my head about the remaining 36 solutions that are not included in my answer. Yes, Positive Integer solutions. Vishwak Srinivasan · 2 years ago

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@Vishwak Srinivasan Even i got 220 on the second try! Adarsh Kumar · 2 years ago

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@Vishwak Srinivasan oh!ok let me see where i have made a mistake! Adarsh Kumar · 2 years ago

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