A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is __________

Solution

Maths sir · 3 steps

Split the 12 into 5 language and 7 other courses, then add the cases with 0, 1 and 2 language courses.

1. Split the pool. There are 5 language courses and 12 - 5 = 7 other courses. At most two language courses means 0, 1 or 2 of them, and the rest of the five come from the 7 others.

   Language courses	Other courses	Ways
   0	5	C(5, 0)C(7, 5) = 1 × 21 = 21
   1	4	C(5, 1)C(7, 4) = 5 × 35 = 175
   2	3	C(5, 2)C(7, 3) = 10 × 35 = 350

2. Add the cases. The three cases cannot overlap, so add them: 21 + 175 + 350 = 546. As a check, C(12, 5) = 792, and the forbidden choices with 3, 4 or 5 language courses number 210 + 35 + 1 = 246, and 792 - 246 = 546.
3. So. The boy can choose his five courses in 546 ways.

The common slip: 525. Leaving out the case with no language course at all gives 175 + 350 = 525, not 546. "At most two" includes zero.

The idea to keep. For "at most" or "at least" conditions, break the choice into cases by how many come from the restricted group, multiply within each case, then add the cases.

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