The possible number of blue marbles is A boy has a collection of blue and green marbles. The number of blue marbles belong to the sets $\{2,3,4,\ldots,13\}$. If two marbles are chosen simultaneously and at random from this collection, then the probability that they have different colour is $\frac{1}{2}$. The possible number of blue marbles is
$$(A)\ 2\hspace{1cm}(B)\ 3\hspace{1cm}(C)\ 6\hspace{1cm}(D)\ 10$$
Let there are $x$ blue and $y$ green marbles in the collection. Since two marbles are chosen chosen simultaneously and at random from this collection, the possible color combinations are blue-green, green-green, blue-blue. But now i am stuck. Please help me.

Suppose, there are $m$ blue and $n$ green marbles.
There are $\binom{m+n}{2}=\frac{(m+n)(m+n-1)}{2}$ ways to choose $2$ marbles.
There are $mn$ ways to choose $2$ marbles with different colors.
The probability of getting two marbles with different colours is therefore
$$\frac{2mn}{(m+n)(m+n-1)}$$
So, the probability is $\frac{1}{2}$, if and only if $$(m+n)(m+n-1)=4mn$$
holds.
$1)\ m=2\ :\ n^2+3n+2=8n$ has no solution in $\mathbb N$
$2)\ m=3\ :\ n^2+5n+6=12n$ has the solutions $1$ and $6$.
$3)\ m=6\ :\ n^2+11n+30=24n$ has the solutions $3$ and $10$.
$4)\ m=10\ :\ n^2+19n+90=40n$ has the solutions $6$ and $15$.
