Given, total students =10 number of groups =3 (i.e. A,B and C ) Each group has atleast one student but group C has atmost 3 students. ∴ There are 3 cases depending on number of students in group C. Case I C has 1 student, then
A
B
]←9 students ∴ Number of ways =10C1×[29−2] Case II C has 2 students, then
A
B
]←8 Students. ∴ Number of ways =10C2×[28−2] Case III C has 3 students, then
A
B
]←7 Students.
∴ Number of ways =10C3×[27−2] ∴ Required number of possibilities =10C1(29−2)+10C2(28−2)+10C3(27−2) =27[10C1×4+10C2×2+10C3] =20−90−240 =128[40+90+120]−350 =(128×250)−350=31650