student-cpt
427 Points
Joined April 2012
5775 is the answer First we give the three groups separate identities A,B,C. For A we have to choose 4 students from 12. This can be done in 12C4 ways i.e. 495 ways. Now we choose 4 students for group B from 8 remaining students. This can be done in 8C4 ways i.e. 70 ways. Therefore total number of ways is 495×70 ways i.e. 34650 ways. But in the original question groups are indistinguishable. Hence we should divide 34650 by 6 (the number of ways 3 things can be placed in 3 places) 34650/6 = 5775. In the second problem the three boys are distinct so the number of ways is 34650. E.g. take a simpler problem 6 students to 3 groups. let the students be denoted by 1,2,3,4,5,6. Now if we allocate them to groups A,B,C then
1st allocation: A:1,2 B:3,4 C:5,6
2nd allocation A:3,4 B:5,6 C:1,2
If we remove the labels A,B,C the above allocations are identical as in they divide the students into same groups. however in the second problem the three boys are distinct and number of ways will be 34650