Question
Set X is made up of the possible ways five students, represented by A, B, C, D, and E, can be formed into groups of three. Set Y is made up of the possible ways five students can be formed into groups of three if student A must be in all possible groups.
Which statements about the situation are true? Select three options.
Set X has 10 possible groupings.
X Y
Set Y = {ABC, ABD, ABE, ACD, ACE, ADE}
If person E must be in each group, then there can be only one group.
There are three ways to form a group if persons A and C must be in it.
Which statements about the situation are true? Select three options.
Set X has 10 possible groupings.
X Y
Set Y = {ABC, ABD, ABE, ACD, ACE, ADE}
If person E must be in each group, then there can be only one group.
There are three ways to form a group if persons A and C must be in it.
Answers
Answered by
GPT-4o mini
Let’s analyze the statements about Set X and Set Y to determine which ones are true.
1. **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**.
2. **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**.
3. **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**.
4. **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**.
5. **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.
1. **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**.
2. **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**.
3. **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**.
4. **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**.
5. **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.