Our Underwater Basket-Weaving club has 12 members. In how many ways can the club members split up into four groups of 3 for a friendly speed basket-weaving competition, if Adi and Noor must be on different teams?

Guest Jun 17, 2023

#1**0 **

There are 1440 ways to split the 12 members of the Underwater Basket-Weaving club into four groups of 3 for a friendly speed basket-weaving competition, if Adi and Noor must be on different teams.

To solve this, we can use the following steps:

First, we can count the number of ways to split the 12 members into four groups of 3, without considering the restriction that Adi and Noor must be on different teams. This can be done using the following formula:

n! / (4! * (n - 12)!)

where n is the total number of members, which is 12.

Plugging this in, we get:

12! / (4! * 8!) = 495

Now, we need to subtract the number of ways to split the members into four groups of 3, with Adi and Noor on the same team. This can be done by counting the number of ways to choose two members from the 10 remaining members, and then dividing by 2, since Adi and Noor can be in either order on the same team.

10! / (2! * 8!) = 45

Finally, we need to add back the number of ways to split the members into four groups of 3, with Adi and Noor on different teams, but in the same group. This can be done by counting the number of ways to choose one member from the 10 remaining members, and then dividing by 2, since Adi and Noor can be in either order in the same group.

10! / (2! * 8!) = 20

Adding these three numbers together, we get the total number of ways to split the members into four groups of 3, with Adi and Noor on different teams:

495 - 45 + 20 = 1440

Guest Jun 17, 2023

#2**0 **

Put Adi in group 1, and Noor in group 2

Choose in 10C2 ways the other members for group 1

.

Choose in 8C2 ways the other members for group 2.

Choose 6C3 ways the members for group 3

Finally we can fill group 4 with the remaining members in 3C3 ways.

Groups 1 and 2 are always unique, groups 3 and 4 are duplicated, so we should be divide by 2!:

**Final result: (10C2*8C2*6C3*3C3) /2! = 12,600 ways.**

Guest Jun 18, 2023