Splitting Up For Study

Using inclusion -exclusion principle,the total number of ways of splitting the students into three groups is $3^8-3(2^8)+3(1^8)$ .However,since the three groups are identical we need to divide by $3!$ . Hence we get the answer as $966$

Relevant wiki: Distinct Objects into Identical Bins

This problem is modeled as distinct objects (students) into identical non-empty bins (study groups).

This can be accomplished using Stirling numbers of the second kind .

In particular, we are trying to find $S(8,3)$ .

Begin with the identity, $S (n,r) = r S(n-1, r) + S(n-1, r-1)$ Use recurrence relations and the basic cases of Stirling numbers of the second kind to find the value of $S(8,3)$ .

$S(8,3)=3S(7,3)+S(7,2)=3S(7,3)+2^6-1=3S(7,3)+63$ $S(7,3)=3S(6,3)+S(6,2)=3S(6,3)+2^5-1=3S(6,3)+31$ $S(6,3)=3S(5,3)+S(5,2)=3S(5,3)+2^4-1=3S(5,3)+15$ $S(5,3)={5\choose 3} + 3{5\choose 4}=10+3\times 5=25$

Now working backwards to $S(8,3)$ :

$S(6,3)=3S(5,3)+15=3\times 25+15=90$ $S(7,3)=3S(6,3)+31=3\times 90+31=301$ $S(8,3)=3S(7,3)+63=3\times 301+63=966$

Thus, there are $\boxed{966}$ ways to form the study groups.