set x is made up of the possible ways five students, represented by a, b, c, d, and e, can be formed into…

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 c 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.

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 c 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.

Answer

Explanation:

Step1: Calculate number of groups in Set X

Use combination formula $C(n,k)=\frac{n!}{k!(n - k)!}$, where $n = 5$ and $k=3$. So $C(5,3)=\frac{5!}{3!(5 - 3)!}=\frac{5!}{3!2!}=\frac{5\times4}{2\times 1}=10$. So Set X has 10 possible groupings.

Step2: Analyze Set Y

Set Y has students with A always in the group of three. We choose 2 students out of the remaining 4. Using combination formula $C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!}=\frac{4\times3}{2\times1}=6$, and the groups are {ABC, ABD, ABE, ACD, ACE, ADE}.

Step3: Analyze if A and C are in the group

If A and C are in the group, we need to choose 1 more student out of the remaining 3. Using combination formula $C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3$.

Step4: Analyze if E is in each group

If E is in each group of three, we choose 2 students out of the remaining 4, $C(4,2) = 6\neq1$.

Step5: Analyze subset relation

Set Y is a subset of Set X, not the other - way around. So $X\not\subset Y$.

Answer:

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.