Suppose you have 20 objects. Three objects are special, and you want them to remain together. You split the objects into two groups of 10 uniformly at random. What is the probability that the three objects remain together?
My work:
In total there are
$$\frac{20!}{10!10!2}$$ ways to separate the objects.
but by wanting 3 on the same team, we can do this in $$\frac{17!}{7!10!2}$$ ways, since we can choose our three special objects in any order, and then must divide the remaining 17.
Is this correct?
Edit
I am being told that it is actually $18$ choose $8$, since if we treat the group of 3 as a single object, then we have 18 objects. Is this true?
To clarify the question, let us say that you and two friends are playing football and want to be on the same team. The teams are split at random, what is the probability that you end up on the same team? Moreover, the teams are split 10 at a time.
We convert this problem to the following:
The probability of drawing out all the special balls is$$P_1=\dfrac{\binom{3}{3}\binom{17}{7}}{\binom{20}{10}}$$and the probability of drawing out no special ball is $$P_2=\dfrac{\binom{3}{0}\binom{17}{10}}{\binom{20}{10}}$$therefore the general probability attains by merging the two cases$$P=P_1+P_2=2\dfrac{\binom{17}{7}}{\binom{20}{10}}$$