Probability of Occupying Adjacent Seats in a Random Arrangement

[JOhnJDC]

Homework Statement

If k people are seated in a random manner in a row containing n seats (n>k), what is the probability that the people will occupy k adjacent seats in the row?

I realize that there are n choose k sets of k seats to be occupied, and that there are n-k+1 sets of k adjacent seats. So the probability that I'm looking for is:

(n-k+1)/(n choose k)

What I don't understand is how the above probability simplifies to:

[(n-k+1)!k!]/n!

Can someone please explain? Thanks.

Homework Equations

(ⁿ_k) = n choose k = n!/[(n-k)!k!]

The Attempt at a Solution

(n-k+1)/(n choose k) = [(n-k+1)(n-k)!k!]/n!

Not sure what to do from here.

 

[rakalakalili]

Look at (n-k+1)(n-k)!, can you simplify this product any?

 

[JOhnJDC]

Other than finding a quotient, I've never had to manipulate/simplify factorials. Thinking about this again, since the number (n-k+1) is 1 greater than the number (n-k), (n-k)! times (n-k+1) must equal (n-k+1)!.