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