There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100 k, then k is equal to _____________.

Solution

Maths sir · 3 steps

Classes 10 and 11 together give only 4 or 5 students, so list those splits and let class 12 fill the rest.

1. Set up. Let x, y, z be the numbers chosen from classes 10, 11 and 12. We need x + y + z = 10, each of x, y, z ≥ 2, and x + y ≤ 5.
2. Find the cases. Since x ≥ 2 and y ≥ 2, we get x + y ≥ 4. With x + y ≤ 5, only x + y = 4 or 5 is possible, so z = 6 or 5. Both are within the 8 students of class 12.

   (x, y, z)	Ways	Count
   (2, 2, 6)	C(5, 2)C(6, 2)C(8, 6) = 10 × 15 × 28	4200
   (2, 3, 5)	C(5, 2)C(6, 3)C(8, 5) = 10 × 20 × 56	11200
   (3, 2, 5)	C(5, 3)C(6, 2)C(8, 5) = 10 × 15 × 56	8400

   4200 + 11200 + 8400 = 23800 = 100 × 238

3. So. The total number of selections is 23800 = 100k, so k = 238.

The common slip: 196. Counting only the cases with exactly 5 from classes 10 and 11 gives 19 600, so k = 196. The split 2 + 2 with 6 from class 12 also satisfies 'at most 5'.

The idea to keep. With several 'at least' and 'at most' conditions, turn them into equations on the counts, list every allowed split, and multiply the choices for each class within each split.

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