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⊂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⊂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 combinations for set X

Use combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 5$ and $r=3$. So $C(5,3)=\frac{5!}{3!(5 - 3)!}=\frac{5!}{3!2!}=\frac{5\times4}{2\times 1}=10$.

Step2: Calculate number of combinations for set Y

Since student A must be in all groups, we need to choose 2 more students out of 4. Using combination formula with $n = 4$ and $r = 2$, $C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4!}{2!2!}=\frac{4\times3}{2\times1}=6$, and set $Y={ABC,ABD,ABE,ACD,ACE,ADE}$.

Step3: Analyze if person E must be in each group

If person E must be in each group, and we are forming groups of 3, choosing 2 more students out of 4 gives $C(4,2) = 6$ groups, not 1.

Step4: Analyze if A and C must be in the group

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

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.