PSL4#63

Here's another approach: what are the odds that the group ronnie is in contains ben and javon? there are \(\binom{23}{2}\) combinations for possible partners for ronnie, one of those options is {ben, javon}, so the odds are \(\frac{1}{\binom{23}{2}}=\frac{1}{253}\)

there are 24C3 ways to choose 3 students from 24

then

21C3 ways to choose 3 from the remaining 21

continuing with the pattern

there are 

24C3 * 21C3 * ....... * 3C3   ways to split 24 kids into 8 ordered groups of 3 BUT our groups are not ordered so I have to divide by 8!

similarly

if we put Ronnie, Ben  and Jevon into 1 group then we are left with 21 kids to put into 7 groups. So we get

21C3 * 18C3 * ....... 3C3     then we must divide by 7!

To the probability that those 3 boys will end up in the same group will be

\(=\frac{21C3 * 18C3 * ....... 3C3 }{7!}\div \frac{24C3 * 21C3 * 18C3 * ....... 3C3 }{8!}\\ =\frac{21C3 * 18C3 * ....... 3C3 }{7!}\times \frac{8!}{24C3 * 21C3 * 18C3 * ....... 3C3 }\\ =\frac{1 }{1}\times \frac{8}{24C3 }\\ =8\div \frac{24!}{3!21!}\\ =8\times \frac{3!21!}{24!}\\ =8\times \frac{6}{22*23*24}\\ =\frac{6}{22*23*3}\\ =\frac{1}{11*23}\\ =\frac{1}{253}\\\)

Melody: I'm no good at this stuff!, but what do think of this naive way:
There are 24C3 =2024 ways of splitting 24 kids.
There are 24/3 =8 ways of splitting them into groups of 3
The probability that Ronnie, Ben and Jevon are in the same group of 3's is: 8 / 2024 = 1 /253