How many socks would you have to draw from the dark room in the worst case, to have exactly $N$ pairs?

Question

There are $A$ red socks, $B$ green socks and $C$ yellow socks in a dark room. $N$ pairs of matching colored socks from the dark room have to be drawn. Given all $A,\ B$, and $C$ are even and $N \leq \frac{1}{2}(A+B+C)$, how many socks would you have to draw from the dark room in the worst case, to have exactly $N$ pairs?

Answer 1

Let's consider a set I1 and an application f1:I1 -->{R,G,Y}. In order to be sure of getting a #f^{-1}(x)=2 for x in {R,G,Y}, at least #I1=#{R,G,Y}+1=4. At least there exists two elements i and j of I1 such that f1(i)!=f1(j). Let's consider the set NC(I1) that contains the elements of I1 which do not form a couple. Let's consider the set I2 such that the application f2:CN(I1)+I2 -->{R,G,Y} satisfies the same properties as f1, in that case at least #[CN(I1)+I2]=4 so #I2=2. And so on, at least we need to draw 4 +2(N-1)=2(N+1) from the dark room to be sure of getting N pairs of socks. In this case we must need A+B+C>=2(N+1).

Answer 2

To generalize the result with p different colour of socks in the dark room we need to draw $p+1+2(N-1) at least from the dark house in order to be sure of getti g N pairsof socks.