How many ways can a 10-player basketball team line up if the 2 captains must stand

JSmith

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Sep 21, 2012
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How many ways can a 10-player basketball team line up if the 2 captains must stand next to each other?

So I know if the captains did not have to stand beside each other, then:

showimage


How do I modify this/work from here to determine if the captains have to stand beside each other how many different ways they could line up?
 
Hello, JSmith!

How many ways can a 10-player basketball team line up
if the 2 captains must stand next to each other?

Let \(\displaystyle A\) and \(\displaystyle B\) be the two captains.
Duct-tape them together.

Now we have nine "people" to arrange.
There are \(\displaystyle 9!\) possible orders.

But for each of them, the two captains
could stand \(\displaystyle AB\) or \(\displaystyle BA\).

Therefore, there are \(\displaystyle 2 \times 9! \,=\,725,\!760\) arrangments.
 
Thanks for your response!!

Why would I not multiply by 8? If there are the two captains and only 10 people, that leaves 8 people we have to place... correct? Or by grouping the captains together create the situation where they only count as one?
Hello, JSmith!


Let \(\displaystyle A\) and \(\displaystyle B\) be the two captains.
Duct-tape them together.

Now we have nine "people" to arrange.
There are \(\displaystyle 9!\) possible orders.

But for each of them, the two captains
could stand \(\displaystyle AB\) or \(\displaystyle BA\).

Therefore, there are \(\displaystyle 2 \times 9! \,=\,725,\!760\) arrangments.
 
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