The students S1,S2,…,S10 are to be divided into 3 groups A,B and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is
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Solution
Number of ways =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 =10(3165) =31650