1. ## Permutations/Combinations

Determine the number of ways that the 12 members of the boys' basketball team can be lined up if Joe, Taner and Josh must all be together?

10! x 2

The answer they have is 604800, which is derived from 10!/3!
How is this a permutations with identical terms? No where in this question does it say that Joe, Taner, and Josh are triplets. What's wrong with my thinking?

2. I'm getting a different answer.

Treat the 3 people who must be together as 1 person. Then there are 10! ways to arrange them. Within this group of 3, there are 3! ways for them to be arranged. so I get 10! x 3!.

3. EDIT:

I meant that I used 10! x 3! also, but they have 10!/3!

4. Hello, skeske1234!

I don't agree with their answer.

Determine the number of ways that the 12 members of the boys' basketball team
can be lined up if $A, B\text{ and }C$ must all be together?

The answer they have is 604,800, which is derived from 10!/3! ??

Duct-tape $A,B,C$ together.
. . Note that there are $3! = 6$ ways they could be ordered.

Then we have 10 "people" to arrange: . $\boxed{ABC}\;D\;E\;F\;G\;H\;I\;J\;K\;L$
. . and there are $10!$ ways.

Therefore, there are: . $10! \times 6$ possible line-ups.

5. Oh man I was sitting here for like 10 minutes trying to figure out just how the hell they managed to get that. I had the exact same thought, "Why the heck would you divide by the permutation of the boys and not multiply!"