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? check all that apply. 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.

If 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, then the set X consists of such triples {ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE} (note that triple ABC is the same as triple ACB, or BCA, or BAC, or CAB, or CBA). The set X totally contains 10 elements (triples). The first statement is true.

If 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, then the set Y consists of such triples {ABC, ABD, ABE, ACD, ACE, ADE} and contains 6 elements. The second statement is also true.


If person E must be in each group, then there can be only one group is false statement, because you can see from the set X that triples which contains E are 6.

There are three ways to form a group if persons A and C must be in it. This statement is true and these groups are ABC, ACD, ACE.

Statements 1 and 3 are true but the 2 statement is false.

Sets

Any collection of elements, which may be mathematically or not.

Given

Five students A, B, C, D, E.

How to calculate the set?

A set X consisting of triple (ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE) is 10.

Hence, the statement first is true.

A set Y is made in which A must be in all groups so our set becomes (ABC, ABD, ABE, ACD, ACE, ADE) is 6.

similarly

A set is made in which E must be there. Condition is similar to the second statement.

Hence, the second statement is false.

A set is made in which A and C must be there (ABC, ACD, ACE).

Hence, the third statement is true.

Thus, statements 1 and 3 are true but the 2 statement is false.

More about sets link is given below.

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