# Permutations/Combinations

• Jan 19th 2010, 07:16 AM
skeske1234
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?
• Jan 19th 2010, 08:07 AM
qmech

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!.
• Jan 19th 2010, 08:13 AM
skeske1234
EDIT:

I meant that I used 10! x 3! also, but they have 10!/3!
• Jan 19th 2010, 08:17 AM
Soroban
Hello, skeske1234!

I don't agree with their answer.

Quote:

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

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

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

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

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

• Jan 19th 2010, 08:33 AM
ANDS!
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!"