Permutations and Combinations Part 1

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
←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
←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