Suppose a number, say $$n$$ is given.

How to find number of solutions of the equation?

$$x \times y \times z = n$$

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, 11 months 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

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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?

- 2 years, 11 months 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 $$4^{4}=256$$

- 2 years, 11 months ago

Aha. Okay. Thanks bro!

- 2 years, 11 months ago

can u help me with this

- 2 years, 11 months ago

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

- 2 years, 11 months ago

can u explain in detail, plz

- 2 years, 11 months ago

$$\lambda_{K_{\alpha}}$$does not change with change in potential difference.

Hence,

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

- 2 years, 11 months ago

Now solve for $$\lambda_{K_{\alpha}}$$.

Then use Moseley's equation.

- 2 years, 11 months ago

thanks for the help

- 2 years, 11 months ago

No problem dude. Hope you got the answer.

- 2 years, 11 months 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.

- 2 years, 11 months ago

- 2 years, 11 months ago

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

- 2 years, 11 months ago

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.

- 2 years, 11 months ago

Even i got 220 on the second try!

- 2 years, 11 months ago

oh!ok let me see where i have made a mistake!

- 2 years, 11 months ago