Alice and Bob are two students among a class of $21$. Their teacher will randomly divide the class into $4$ groups of $4$ and one group of $5.$ Find the probability that Alice and Bob are put in the same group?
A possible way of solving this problem outlined is to do $(4)(4)(3)$ and $(5)(4),$ add them and divide by $420.$ However, I'm not really sure how this works. If we split the problem into cases of when they are together in a group of $4$ and when they are together in a group of $4,$ for the first case there are $4$ possible groups, but from there why do we have to do $4 \cdot 3?$ If we want the two to be in the same group, then there will be $19$ undetermined spots, and of those we can choose $2$ people to be in that group of $4,$ to obtain $\tbinom{19}{2} \cdot 4=15504,$ but this is clearly wrong. What am I doing wrong?
Also, I was wondering if there is a easier way of doing this problem instead of computing number of desired possibilities over total possibilities, and instead considering the problem as, since there are $4$ groups of $4,$ then there is a $\tfrac{16}{21}$ possibility of groups of $4,$ and a $\tfrac{5}{21}$ possibility of groups of $5.$ In other words, I am curious if there are any other approaches to the aforementioned one, which I am confused on why it works. Thanks in advance!
Best Answer
Rare problem that is best attacked as probability of events problem rather than enumeration of Combinatorics.
Probablity that Alice is at a table of $(4)$ is $\dfrac{16}{21}.$
When that happens, probability that Bob is at the same table is $\dfrac{3}{20}.$
Probablity that Alice is at a table of $(5)$ is $\dfrac{5}{21}.$
When that happens, probability that Bob is at the same table is $\dfrac{4}{20}.$
Putting this all together, the desired computation is
$$\left[\frac{16}{21} \times \frac{3}{20}\right] + \left[\frac{5}{21} \times \frac{4}{20}\right].$$