Let’s analyze the statements about Set X and Set Y to determine which ones are true.
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Set X has 10 possible groupings.
- To find the number of ways to choose 3 students from 5 (students A, B, C, D, E), we use combinations: \[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
- This statement is true.
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X < Y
- Set Y includes combinations where A must be in every group of three. From the remaining students (B, C, D, E), we choose 2: \[ \binom{4}{2} = 6 \]
- Set Y consists of the following groups: ABC, ABD, ABE, ACD, ACE, ADE. Set Y has 6 combinations, while Set X has 10. Thus, Set X is greater than Set Y. This statement is false.
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Set Y = {ABC, ABD, ABE, ACD, ACE, ADE}
- As previously calculated, Set Y does indeed consist of groups {ABC, ABD, ABE, ACD, ACE, ADE}. This statement is true.
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If person E must be in each group, then there can be only one group.
- If E is in the group, we need to choose 2 more from A, B, C, D: \[ \binom{4}{2} = 6 \]
- The possible groups would be {ABE, ACE, ADE, BEC, CED, etc.} So there can be multiple groups. This statement is false.
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There are three ways to form a group if persons A and C must be in it.
- If A and C are already selected, we need to choose 1 more from B, D, or E: \[ \binom{3}{1} = 3 \]
- The groups formed would be {ABC, ACD, ACE}. This statement is true.
Based on the analysis above, the three true statements about the situation are:
- Set X has 10 possible groupings.
- Set Y = {ABC, ABD, ABE, ACD, ACE, ADE}.
- There are three ways to form a group if persons A and C must be in it.
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