For this we have to find number of ways we can get a sum of 10 in 3 rolls of dice which is nothing but the number of positive integral solutions of the equation :
x1 + x2 + x3 = 10 where 1 <= x1,x2,x3 <= 6
Now number of possible integral solutions can be found as :
Coefficient of x^{10} in (x + x^{2} + ........ + x^{6})^{3}
==> Coefficient of x^{10} in x^{3} (1 + x +.....+ x^{5})^{3}
==> Coefficient of x^{7} in (1 + x +.....+ x^{5})^{3}
==> Coefficient of x^{7} in (1 - x^{6})^{3} . (1 - x)^{-3}
Now only relevant terms will be extracted from (1 - x^{6})^{3} and accordingly powers of (1 - x)^{-3 }will be taken to get the coefficients of x^{7} .
So (1 - x^{6})^{3 }= (1 - 3x^{6} + 3x^{12} - x^{18}) out of which only first 2 terms are relevant..
So , due to the term "1" , coefficient of x^{7} is needed in (1 - x)^{-3} which is : ^{3-1+7}C_{7} = ^{9}C_{7} = 36
due to term "x^{6 }" , coefficient of x is needed in (1 - x)^{-3} which is : ^{3-1+1}C_{1} = ^{3}C_{1}_{ }= 3
So due to -3x^{6} , we will have coefficient = -3 * 3 = -9
Hence net coefficient of x^{7} = 36 - 9 = 27
Hence P(sum of 3 dices is 10) = n(sum of 3 dices is10) / Total outcome
= 27 / (6 * 6 * 6)
= 27 / 216
= 1 / 8