First, lets assume that a,b,c are all +ve to simplify the question.
The three numbers can have the following combinations:
118 --> with 3 rearrangements ( because 1 is repeated, the possibilities = \(\frac{3!}{2!} = 3\))
127 --> with 6 rearrangements (because all numbers are different, the possibilities = \(3! = 6\))
136 --> with 6 rearrangements
145 --> with 6 rearrangements
226 --> with 3 rearrangements
235 --> with 6 rearrangements
244 --> with 3 rearrangements
334 --> with 3 rearrangements

Total = 36
(the above part is similar to another interesting counting question mentioned here: (https://gmatclub.com/forum/in-how-many-ways-10-identical-chocolates-be-distributed-among-3-child-307384.html#p2375066)

However, each of a,b,c can be either +ve or -ve,
so the combined possibilities of the signs = 2*2*2 = 8
for details, (a,b,c) can be:
(+,+,+)
(+,+,-)
(+,-,+)
(-,+,+)
(+,-,-)
(-,+,-)
(-,-,+)
(-,-,-)
8 possibilities

so the total number of possibilities = 36*8 = 288
D

|a|+|b|+|c|=10

integral solutions possible for (|a|, |b|, |c|)= 9C2=36

Now a, b and c can can be positive or negative; Hence total possible solutions for (a,b,c)= 2*2*2*36= 288

Hello, could you please explain why you're doing 9C2 instead of 9C3?

Because the formula is n-1cr-1

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Whats the reasoning behind 9c2, can you please explain?

Whats the reasoning behind 9c2, can you please explain?

Assume 3 dies first, that add up to 10. Each has numbers from 1 - 6, and the no. of ways to get a sum of 10 on 3 dies = 27 ways

Each no. can be +ve or -ve, so each number can be selected in 2 ways.

3 No.s, each + or -ve adding to 10 = 27 * 2 * 2 * 2 = 216 ways

There will be ONLY two additional possibilities: i.e. (7,2,1) and (8,1,1)

So including this, will increase the total from 216 but definitely not up to 576.

D

PS: Exact numbers can be calculated, but not needed at this point:
(7,2,1) and (8,1,1) can be re-arranged in : 3! + (3!)/2 ways = 9
each can be positive or negative, so total as above: 9 * 8 = 72
Total solutions = 216 + 72 = 288