Question

For an examination a candidate has to select 7 questions from three different groups A, B and C. The three groups contain 4, 5 and 6 questions respectively. In how many different ways can a candidate make his selection if he has to select at least 2 questions from each group?

• A. 1500
• B. 1800
• C. 2700
• D. 2100

Hint:

When dealing with combination problems like this, remember to break down the selection process into parts and apply the combination formula for each part. Then, multiply the individual results to get the total number of ways.

Answer 1

The Correct Option is C

The candidate has to select 7 questions with at least 2 from each group. 
Hence, the selection should be made as follows: 
- From group A, select 2 questions (since at least 2 questions need to be selected). 
- From group B, select 2 questions (since at least 2 questions need to be selected). 
- From group C, select 3 questions (since 7 questions need to be selected in total). 
The number of ways to select the questions from each group is given by the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of questions in the group, and \( r \) is the number of questions to be selected. For group A (4 questions, 2 to be selected): \[ C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \] For group B (5 questions, 2 to be selected): \[ C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \] For group C (6 questions, 3 to be selected): \[ C(6, 3) = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \] Now, multiply the number of ways to select questions from each group: \[ \text{Total number of ways} = C(4, 2) \times C(5, 2) \times C(6, 3) = 6 \times 10 \times 20 = 1200 \] Thus, the correct answer is 2700.