Let's analyze each given statement one by one with detailed explanations:
1. Set [tex]$X$[/tex] has 10 possible groupings:
To determine the number of ways to form groups of 3 out of 5 students, we use combinations. The formula for combinations is given by:
[tex]\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}
\][/tex]
Here, [tex]\( n = 5 \)[/tex] and [tex]\( k = 3 \)[/tex]. Thus:
[tex]\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = 10
\][/tex]
So, Set [tex]$X$[/tex] indeed has 10 possible groupings.
2. [tex]$x \subset y$[/tex]:
Set [tex]$X$[/tex] is the set of all possible groups of 3 students from 5 students. Set [tex]$Y$[/tex] is the set of all possible groups of 3 students from 5 students, with the condition that student [tex]$A$[/tex] must be in every group. By definition, [tex]$Y$[/tex] is a subset of [tex]$X$[/tex] because every group in [tex]$Y$[/tex] must also be a valid group in [tex]$X$[/tex]. However, not all groups in [tex]$X$[/tex] are included in [tex]$Y$[/tex] since [tex]$A$[/tex] may not necessarily be in them. Therefore, this statement should be written as [tex]$Y \subset X$[/tex], not [tex]$X \subset Y$[/tex]. So, this statement is false.
3. Set [tex]$Y$[/tex] is \{ABC, ABD, ABE, ACD, ACE, ADE\}:
To form groups of 3 students with student [tex]$A$[/tex] included in every group, we choose 2 more students from the remaining 4 students [tex]\( (B, C, D, E) \)[/tex].
The combinations for selecting 2 out of these 4 are:
[tex]\[
\binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6
\][/tex]
Hence, the groups are:
[tex]\[
ABC, ABD, ABE, ACD, ACE, ADE
\][/tex]
So, this statement is true.
4. If person [tex]$E$[/tex] must be in each group, then there can be only one group:
If student [tex]$E$[/tex] must be in each group of 3, we are left to choose 2 more students from the remaining 4 students [tex]\( (A, B, C, D) \)[/tex].
The combinations are:
[tex]\[
\binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6
\][/tex]
Hence, the groups are:
[tex]\[
EAB, EAC, EAD, EBC, EBD, ECD
\][/tex]
Clearly, there are 6 different groups possible, not just one. So, this statement is false.
5. There are three ways to form a group if persons [tex]$A$[/tex] and [tex]$C$[/tex] must be in it:
If both students [tex]$A$[/tex] and [tex]$C$[/tex] must be in the group, we need to choose 1 more student from the remaining 3 students [tex]\( (B, D, E) \)[/tex].
The combinations are:
[tex]\[
\binom{3}{1} = \frac{3!}{1! \cdot 2!} = \frac{6}{2} = 3
\][/tex]
Hence, the groups are:
[tex]\[
ABC, ACD, ACE
\][/tex]
So, this statement is true.
In conclusion, the true statements about the situation are:
- Set [tex]$X$[/tex] has 10 possible groupings.
- Set [tex]$Y = \{ABC, ABD, ABE, ACD, ACE, ADE\}$[/tex].
- There are three ways to form a group if persons [tex]$A$[/tex] and [tex]$C$[/tex] must be in it.
