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Question

There are 10 chairs in a row. In how many ways can three people be seated and in how many of these will the two end chairs be occupied.

Answer 1

There are 10 choices for seating the first person, times 9 choices for the second person, times 8 for the third. It's a common combinatorial situation, where you use the notation (can't write it in here properly) "10 choose 3."

(10 3) (but the 3 is written underneath the 10 with no fraction bar)

10 x 9 x 8 = 720.

Now, we see how many combinations there are with the two ends occupied. There are three choices for the person in chair 1, times two choices for the person in chair 10, times 8 choices for where the third person goes.

3 x 2 x 8 = 48 arrangements of this type.

Edit: Puzzling, aren't we to assume that Bob in chair 1, Joe in chair 2, Lou in chair 7 is not the same as Joe in 1, Bob in 2, and Lou in 7? This seemed like an "order is significant" type of question to me.

Asker: If you're only concerned with the empty/taken state of the chairs, and not who sits in them, then you need to divide out the re-arrangements of the people like Puzzling said.

Answer 2

PART A:

There are 10 choose 3 ways to seat the people in the chairs.

C(10, 3) = 10 x 9 x 8 / 3!

= 720 / 6

= 120 ways

(If you are unsure, just imagine that there are 10 ways to place the first person, 9 for the second and 8 for the third.  However, you end up overcounting the arrangements by a factor of 6 because you can arrange the 3 people into the same chairs 3! = 6 ways)

Answer:

120 ways

PART B:

Now imagine putting two people in the end chairs.  You have 8 ways to place the 3rd person, so there are only 8 of these that both end chairs are occupied.

Answer:

8 ways