To determine how many ways we can select 2 students from a group of 5 students, we can use the concept of combinations. Combinations are used when the order of selection does not matter, which is the case here since we are simply selecting students for a debating competition. ### Step-by-Step Explanation 1. **Understanding Combinations**: - When we select items from a group and the order does not matter, we use combinations. The formula for combinations is given by: \[ C(n, r) = \frac{n!}{r!(n - r)!} \] - Here, \( n \) is the total number of items (students, in this case), \( r \) is the number of items to choose, and \( ! \) denotes factorial, which is the product of all positive integers up to that number. 2. **Identifying Values**: - In our problem, we have: - \( n = 5 \) (the total number of students) - \( r = 2 \) (the number of students we want to select) 3. **Applying the Formula**: - Plugging the values into the combinations formula: \[ C(5, 2) = \frac{5!}{2!(5 - 2)!} = \frac{5!}{2! \cdot 3!} \] 4. **Calculating Factorials**: - Now, we need to calculate the factorials: - \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) - \( 2! = 2 \times 1 = 2 \) - \( 3! = 3 \times 2 \times 1 = 6 \) 5. **Substituting Back**: - Now substitute these values back into the formula: \[ C(5, 2) = \frac{120}{2 \times 6} = \frac{120}{12} = 10 \] 6. **Final Answer**: - Therefore, the number of ways to select 2 students from a group of 5 is **10 ways**. ### Explanation of Other Options - **Option A: 25 ways**: This option might arise from a misunderstanding of the problem, possibly thinking that order matters (which would lead to permutations). However, since we are only interested in combinations, this option is incorrect. - **Option C: 15 ways**: This option does not correspond to any standard calculation for combinations of 5 taken 2 at a time. It may be a miscalculation or confusion with another problem. - **Option D: 20 ways**: Similar to option C, this does not match the correct calculation for combinations. It could stem from an incorrect application of the formula or misunderstanding of the problem. ### Common Pitfalls - **Confusing Combinations with Permutations**: Remember that combinations are used when the order does not matter, while permutations are used when the order does matter. - **Miscalculating Factorials**: Ensure that you calculate factorials correctly, as small errors can lead to incorrect answers. - **Not Using the Correct Formula**: Always double-check that you are using the combinations formula when the order of selection is not important. ### Revision Summary - Use the combinations formula \( C(n, r) = \frac{n!}{r!(n - r)!} \) for selecting items where order does not matter. - For selecting 2 students from 5, the calculation yields \( C(5, 2) = 10 \). - Be cautious of confusing combinations with permutations and ensure correct factorial calculations. - Always verify your understanding of the problem to avoid miscalculations.