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 ___________.

Solution

Maths sir · 3 steps

Fix how many go to C first, then split everyone else between A and B with neither left empty.

1. Case on group C. Group C must have 1, 2 or 3 students. If C has k students, choose them in C(10, k) ways.
2. Split the rest. Each of the remaining 10-k students goes to A or B, which gives 2¹⁰⁻ᵏ ways. Subtract the 2 ways where everyone goes to A or everyone goes to B, so both A and B get at least one student.

   N = Σₖ₌₁³ C(10, k)(2¹⁰⁻ᵏ - 2)

   Students in C	Choose C	Split rest into A, B	Ways
   k = 1	C(10, 1) = 10	2⁹ - 2 = 510	5100
   k = 2	C(10, 2) = 45	2⁸ - 2 = 254	11430
   k = 3	C(10, 3) = 120	2⁷ - 2 = 126	15120

3. So. N = 5100 + 11430 + 15120 = 31650.

The common slip: 32000. Forgetting to subtract the 2 cases where A or B ends up empty gives 5120 + 11520 + 15360 = 32000, not 31650.

The idea to keep. When one group has a size limit, case on that group's size. Then each remaining person picks between the other groups, and you remove the cases that leave a group empty.

Answer from the NTA final answer key. Solution by PrepWiser.