When order does not matter, the number of ways to choose r objects from n is given by the combination formula nCr. Here n is 5 and r is 2, so the number of committees is 5C2. Using the formula, 5C2 = 5! ÷ (2! × 3!) = (5 × 4) ÷ (2 × 1) = 10. Therefore, there are 10 distinct ways to select a pair of students from the group.
Option A:
Option A, 5, corresponds to the number of students rather than the number of pairs that can be formed. It ignores the multiple distinct combinations that are possible when selecting two members.
Option B:
Option B, 8, does not match the combinatorial calculation for 5C2 and appears to be an arbitrary number without basis in the formula.
Option C:
Option C, 9, is close to 10 but still does not equal the true combination count as derived from factorial operations, showing a small but important error.
Option D:
Option D correctly employs the combination formula and simplifies it accurately. It recognizes that for each choice of two students, the order is irrelevant, leading to exactly 10 possible committees.
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