18 people are being randomly selected into groups of 3. Bob and Alice want to know the chance that they would be selected to be in the group together. My first thought was to find all groups where Alice and Bob are together, and then divide that by the total number of combinations of groups. I think this idea is correct, but I'm still a little unsure. This gives 1.96% by the formula
$$\frac{n-p}{\binom{n}{k}}$$
Where $n=18$, $p=2$ ( the number of people you want to be in the same group ) and $k=3$. Can anyone verify whether my solution is correct?
Best Answer
I agree that Lulu's comment represents (by far) the easiest approach to the problem.
I also agree with econbernardo's comment that proposed solutions to problems of this type can be sanity-checked. That is, you will need to be able to write simple computer programs in a language like C, Java, or Python. If you have no programming experience, then I suggest making Python your first programming language.
The (inferior) approach, that involves Combinatorics, is given below.
The probability will be expressed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where
$$D = \binom{18}{3} \times \binom{15}{3} \times \binom{12}{3} \times \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3}.$$
Let $E$ denote the computation
$$\binom{15}{3} \times \binom{12}{3} \times \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3}.$$
Then
$$D = \binom{18}{3} \times E. \tag1 $$
Notice that for convenience, the denominator $D$ was computed so as to assume that people were assigned to group-1, then people were assigned to group-2, and so forth.
This means that $D$ was computed so as to distinguish between a specific group of $3$ people being assigned to group-1, and (for example) the same group of $3$ people being assigned to group-2.
Because this (convenient) method of enumeration was chosen for $D$, a consistent method of enumeration must be chosen for $N$. This means (for example) that the situation where Alice and Bob are assigned to group-1 must be distinguished from the situation where Alice and Bob are assigned to group-2.
The number of satisfying groupings, that have Alice and Bob both assigned to group-1 is
$$\binom{16}{1} \times \binom{15}{3} \times \binom{12}{3} \times \binom{9}{3} \times \binom{6}{3} \times \binom{3}{3}$$
$$= \binom{16}{1} \times E. \tag2 $$
As discussed, the computation of $N$ must distinguish between Alice and Bob being sent to group-1, and Alice and Bob being sent to any of the other groups. By symmetry, the number of assignments where Alice and Bob are assigned to group-2 (for example) is the same as the number of assignments where Alice and Bob are assigned to group-1.
Therefore, using (2) above as a starting point, you have that
$$N = \binom{6}{1} \times \binom{16}{1} \times E = 96 \times E. \tag3 $$.
Therefore, using (1) and (3) above, you have that
$$\frac{N}{D} = \frac{96 \times E}{\binom{18}{3} \times E} = \frac{96}{\binom{18}{3}}$$
$$= \frac{96 \times (3!) \times [(15)!]}{(18)!} = [96 \times 6] \times \frac{(15)!}{(18)!} = \frac{96 \times 6}{18 \times 17 \times 16} = \frac{576}{288} \times \frac{1}{17} = \frac{2}{17}.$$