Permutations and Combinations

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Stardust1

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A minibus has 6 forward-facing and 2 backward-facing seats. If 8 people use the bus, in how many ways can they be seated:
a) with no restrictions
b) if one person must sit in a forward-facing seat
c) if 2 people must sit in a forward-facing seat?
Hi everyone, I'm having trouble solving this question.I can do a) but not b) or c).
 
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Stardust1 said:
A minibus has 6 forward-facing and 2 backward-facing seats. If 8 people use the bus, in how many ways can they be seated:
a) with no restrictions
b) if one person must sit in a forward-facing seat
c) if 2 people must sit in a forward-facing seat?
Hi everyone, I'm having trouble solving this question.I can do a) but not b) or c).
Click to expand...
Please show us your work on (a), and some sort of attempt at (b) or (c), so we can see what sort of trouble you're having. That makes a difference in how we help.

For (b), for example, I might start by modeling the situation as putting ABCDEFGH into [forward:] _ _ _ _ _ _ [backward:] _ _, with A having to be forward-facing.

One arrangement would be H D A F G C E B. How many ways are there to do that?

I might start by (conceptually) choosing a seat for the person who has to be forward-facing, and then fill in the rest.
 
Hi, (a) would be 8! = 40320. However, I've tried solving for (b) by doing 1 x 7! or 2! x 7! or 2! x 6! but the answer is 30240. For (c), I tried 2! x 6! but it doesn't work?
 
Stardust1 said:
Hi, (a) would be 8! = 40320.
Click to expand...
Correct; we are permuting 8 items.
Stardust1 said:
However, I've tried solving for (b) by doing 1 x 7! or 2! x 7! or 2! x 6! but the answer is 30240.
Click to expand...
You didn't say before that you thought you were wrong because you were given a supposedly correct answer. Now we have a better sense of what you meant by "having trouble". (The given answer is correct; that isn't always true.)

But why did you do those things? The goal in solving a problem is not just to randomly get the answer you are supposed to get; you have to convince yourself your answer is right before you check against a given answer. (Life doesn't always give you answers.)

Try using my hint. How many ways are there to place person A? Then, how many ways are there to place the other 7 people?
 
There are 2 ways to place person A - 2!. Then there would be 7 ways to place the other 7 people - 7!. So would it be 2! x 7! then?
 
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