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Okay, so we are given a few concentric circles, we don't know how many. I'll give a few ideas of mine. Let me assume that they aren't equidistant.

Plot the given figure in the Cartesian plane, with the common center as the origin (0,0).

[ A co-ordinate geometric approach]

Now we can fix a point on the first circle, then we can plot the other points variably, and then use the shoelace formula to find the co-ordinates as so to maximise the area.

 

[A]

Firstly, I assume that the question you asking is to find the number of 10-digit numbers whose digit sum is 2,3, and 4 respectively, in separate cases.

Subdivision (a) :- Digit sum is 2

We have that 2 = 0+2 or 1+1.

So we can either have a 10-digit number composed of only one 2 and nine 0's, or a 10-digit number composed of two 1's and eight 0's.

In the first case, there is only one such number, that is 2000000000, and nothing else, because 2 can only be in the first digit's place.

In the latter, we have a 1 to be in the first digit's place and another 1 has 9 other places.

So totally, in this case we have 1 x 9 = 9 such numbers, and in the first case we have $1$ number, so totally there are 10 such numbers.

 

Similarly, you can follow this principle for the other two subdivisions too. Hope this helps.

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See here:

https://artofproblemsolving.com/community/q1h1564035p9576025

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Hint :

You can try to use the identity that $$\binom{n}{0} + binom{n}{1} + ...... + binom{n}{n} = 2^n $$ and then manipulate the rest.

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