Answer :
Let's analyze the situation and each of the statements step by step to determine which statements are true.
### Set [tex]\( X \)[/tex]
Given five students [tex]\( A, B, C, D, \)[/tex] and [tex]\( E \)[/tex], we need to find the number of ways to form groups of three students. The number of possible combinations for forming groups of three from five can be calculated using the combination formula [tex]\( \binom{n}{k} \)[/tex], where [tex]\( n \)[/tex] is the total number of items, and [tex]\( k \)[/tex] is the number of items to choose.
[tex]\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10 \][/tex]
Thus, Set [tex]\( X \)[/tex] has 10 possible groupings.
### Set [tex]\( Y \)[/tex]
Set [tex]\( Y \)[/tex] includes all possible combinations of three students, but student [tex]\( A \)[/tex] must be in every group. To determine the number of ways to form such groups, we note that since [tex]\( A \)[/tex] is already included, we need to choose 2 more students from the remaining 4 ( [tex]\( B, C, D, \)[/tex] and [tex]\( E \)[/tex] ).
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3!}{2! \times 2!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]
Thus, Set [tex]\( Y = \{(A, B, C), (A, B, D), (A, B, E), (A, C, D), (A, C, E), (A, D, E) \}\)[/tex].
### Analyzing the Statements:
1. Set [tex]\( X \)[/tex] has 10 possible groupings.
- True. As calculated above, there are 10 ways to form groups of 3 students from 5.
2. [tex]\( X \subset Y \)[/tex]
- False. Set [tex]\( X \)[/tex] includes all combinations, while set [tex]\( Y \)[/tex] only includes combinations with [tex]\( A \)[/tex]. Therefore, Set [tex]\( X \)[/tex] is not a subset of Set [tex]\( Y \)[/tex].
3. Set [tex]\( Y = \{(A, B, C), (A, B, D), (A, B, E), (A, C, D), (A, C, E), (A, D, E) \} - True. As calculated, Set \( Y \)[/tex] includes these exact combinations.
4. If person [tex]\( E \)[/tex] must be in each group, then there can be only one group.
- False. We can form multiple groups with [tex]\( E \)[/tex] as long as we include other combinations from [tex]\( X \)[/tex]. There would be 6 such combinations: [tex]\((A, B, E), (A, C, E), (A, D, E), (B, C, E), (B, D, E), (C, D, E)\)[/tex].
5. There are three ways to form a group if persons [tex]\( A \)[/tex] and [tex]\( C \)[/tex] must be in it.
- True. The remaining one student can be chosen from [tex]\( B, D, \)[/tex] and [tex]\( E \)[/tex]. Therefore, the groups are [tex]\((A, C, B), (A, C, D), \)[/tex] and [tex]\((A, C, E)\)[/tex].
### Conclusion:
The true statements are:
1. Set [tex]\( X \)[/tex] has 10 possible groupings.
2. Set [tex]\( Y = \{ (A, B, C), (A, B, D), (A, B, E), (A, C, D), (A, C, E), (A, D, E) \} 3. There are three ways to form a group if persons \( A \)[/tex] and [tex]\( C \)[/tex] must be in it.
### Set [tex]\( X \)[/tex]
Given five students [tex]\( A, B, C, D, \)[/tex] and [tex]\( E \)[/tex], we need to find the number of ways to form groups of three students. The number of possible combinations for forming groups of three from five can be calculated using the combination formula [tex]\( \binom{n}{k} \)[/tex], where [tex]\( n \)[/tex] is the total number of items, and [tex]\( k \)[/tex] is the number of items to choose.
[tex]\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10 \][/tex]
Thus, Set [tex]\( X \)[/tex] has 10 possible groupings.
### Set [tex]\( Y \)[/tex]
Set [tex]\( Y \)[/tex] includes all possible combinations of three students, but student [tex]\( A \)[/tex] must be in every group. To determine the number of ways to form such groups, we note that since [tex]\( A \)[/tex] is already included, we need to choose 2 more students from the remaining 4 ( [tex]\( B, C, D, \)[/tex] and [tex]\( E \)[/tex] ).
[tex]\[ \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3!}{2! \times 2!} = \frac{4 \times 3}{2 \times 1} = 6 \][/tex]
Thus, Set [tex]\( Y = \{(A, B, C), (A, B, D), (A, B, E), (A, C, D), (A, C, E), (A, D, E) \}\)[/tex].
### Analyzing the Statements:
1. Set [tex]\( X \)[/tex] has 10 possible groupings.
- True. As calculated above, there are 10 ways to form groups of 3 students from 5.
2. [tex]\( X \subset Y \)[/tex]
- False. Set [tex]\( X \)[/tex] includes all combinations, while set [tex]\( Y \)[/tex] only includes combinations with [tex]\( A \)[/tex]. Therefore, Set [tex]\( X \)[/tex] is not a subset of Set [tex]\( Y \)[/tex].
3. Set [tex]\( Y = \{(A, B, C), (A, B, D), (A, B, E), (A, C, D), (A, C, E), (A, D, E) \} - True. As calculated, Set \( Y \)[/tex] includes these exact combinations.
4. If person [tex]\( E \)[/tex] must be in each group, then there can be only one group.
- False. We can form multiple groups with [tex]\( E \)[/tex] as long as we include other combinations from [tex]\( X \)[/tex]. There would be 6 such combinations: [tex]\((A, B, E), (A, C, E), (A, D, E), (B, C, E), (B, D, E), (C, D, E)\)[/tex].
5. There are three ways to form a group if persons [tex]\( A \)[/tex] and [tex]\( C \)[/tex] must be in it.
- True. The remaining one student can be chosen from [tex]\( B, D, \)[/tex] and [tex]\( E \)[/tex]. Therefore, the groups are [tex]\((A, C, B), (A, C, D), \)[/tex] and [tex]\((A, C, E)\)[/tex].
### Conclusion:
The true statements are:
1. Set [tex]\( X \)[/tex] has 10 possible groupings.
2. Set [tex]\( Y = \{ (A, B, C), (A, B, D), (A, B, E), (A, C, D), (A, C, E), (A, D, E) \} 3. There are three ways to form a group if persons \( A \)[/tex] and [tex]\( C \)[/tex] must be in it.
