Krdytkyu

How can I solve problems like this:

Let ${c_1,c_2,c_3,c_4,c_5,c_6}$ be a random sequence where $c_iin (0,1,2,3,4,5,6,7,8,9)$ What is probability that $c_1+c_2+c_3=c_4+c_5+c_6$,

$c_1$ to $c_6$ is not a number there can be all zeroes and combination like $1,1,1$ etc. is possible.

I have a problem with getting how many is combinations there are. I saw that for a specific number like $c_1+c_2+c_3=12$, it is possible to use a generating function, but I don't know how to use it when we have statement like in this problem.

It seems that you know how to compute $$p_k:=Pbigl[c_1+c_2+c_3=kbigr]qquad(0leq kleq27) .$$
The final result $p$ is then simply given by
$$p=sum_{k=0}^{27}p_k^2 .$$
By the way: The generating function for the number of choices of $c_1$, $c_2$, $c_3$ summing to a given $k$ is
$$left(sum_{i=0}^9 x^iright)^3=(1-x^{10})^3(1-x)^{-3}=(1-3x^{10}+3x^{20}-x^{30}bigr)sum_{j=0}^infty{2+jchoose j}x^j .$$

You are looking for
$$N_{,b} (s,r,m) = text{No}text{. of solutions to};left{ begin{gathered}
0 leqslant text{integer }x_{,j} leqslant r hfill \
x_{,1} + x_{,2} + cdots + x_{,m} = s hfill \
end{gathered} right.$$

which is given by
$$
N_b (s,r,m)quad left| {;0 leqslant text{integers }s,m,r} right.quad =
sumlimits_{left( {0, leqslant } right),,k,,left( { leqslant ,frac{s}{r+1}, leqslant ,m} right)}
{left( { - 1} right)^k binom{m}{k}
binom
{ s + m - 1 - kleft( {r + 1} right) }
{ s - kleft( {r + 1} right)} }
$$
as widely explained in this related post.

Of course in this case it is $m=3$ , $r=9$ amd $0 le s le 27$.