Originally Posted by
awkward
Hi Milagros,
Soroban has given you an excellent solution already, but here is a generating function approach since that seems to be what you are looking for.
First, let
$\displaystyle a_r$ = the number of ways to form a binary string of length 50 with exactly r ones, and let
$\displaystyle f(x) = \sum a^r x^r$.
We know
$\displaystyle a_r = \binom{50}{r}$,
so by the binomial theorem,
$\displaystyle f(x) = (1 + x)^{50}$.
Now let
$\displaystyle b_r$ = the number of ways to form a binary string of length 50 with exactly r ones, all at the end. There is only one such string, so
$\displaystyle b_r = 1$
and
$\displaystyle g(x) = \sum b_r x^r = 1 + x + x^2 + \dots + x^{50} = \frac{1-x^{51}}{1-x}$
Now the number of ways to form a binary string of length 50 and another string of length 50 with all the one's at the end, containing a total of r ones, is
$\displaystyle \sum a_r b_{50-r}$
which has the generating function $\displaystyle f(x) \cdot g(x)$.
So the answer to the original question is the coefficient of $\displaystyle x^{10}$
in
$\displaystyle \frac{(1+x)^{50} \cdot (1-x^{51})}{1-x}$
which I think probably reduces, in the end, to the same answer given by Soroban (but I haven't checked it).