Final answer:
The club can select a team in 165 ways with Mary, 330 ways without either twin, and 385 ways with at least one twin on the team.
Step-by-step explanation:
The computing club must select a team of four to represent the club in a hackathon, and the scenarios involve the participation of twins Mary and Anna.
a) Mary is definitely on the team
If we want to select a team with Mary, we already have one member selected, leaving us with three spots to fill. Since we cannot select Anna because she is a twin (as Mary's presence implies Anna's absence), we have 11 other members to choose from.
The number of ways to select the remaining three members is given by the combination formula, which is C(n, r) = n! / (r! * (n - r)!), where n is the total number of available members, and r is the number of members to select.
With n=11 and r=3, we'll have C(11, 3) ways to form the team.
Combination calculation: C(11, 3) = 11! / (3! * (11 - 3)!)
= 11! / (3! * 8!)
= (11 * 10 * 9) / (3 * 2 * 1)
= 165 ways.
b) Neither of the twins can participate
If neither twin can participate, we have 11 (13 total members minus the twins) potential members to select the team of four. Using the combination formula C(11, 4), we calculate how many teams can be formed.
Combination calculation: C(11, 4) = 11! / (4! * (11 - 4)!)
= 11! / (4! * 7!)
= (11 * 10 * 9 * 8) / (4 * 3 * 2 * 1)
= 330 ways.
c) At least one of the twins can participate
If at least one twin must be on the team, we remove the scenario where neither participates and account for the ways where we have either Mary or Anna, but not both, on the team. We already calculated the teams without any twins as 330, so we subtract this from the total possible teams selecting any four from the 13 members, which is C(13, 4), and then calculate this value.
Combination calculation: C(13, 4) = 13! / (4! * (13 - 4)!)
= 13! / (4! * 9!)
= (13 * 12 * 11 * 10) / (4 * 3 * 2 * 1)
= 715 ways to select any team of four.
Now subtract the 330 ways without twins: 715 - 330 = 385 ways where at least one twin can participate.