Asked by Joey

Can someone help me solve this question? I've read the book, notes, I still don't get it...

Simply each of the following.

a) n!/(n-1)!
b) (n-r)!/(n-r-1)!
Answered by Reiny
think of numbers, eg 6!
what is 6!/5! ?

= 6x5x4x3x2x1/(5x4x3x2x1)
= 6

same way for n!/(n-1)!
= n(n-1)(n-2)..(3)(2)(1) / [(n-1)(n-2)..(3)(2)(1)]
= n

try the other problem
Answered by Joey
I don't understand it...

n(n-r)(n-r-1)(n-r-2)..(3)(2)(1)/(n-r-1)(n-r-2)..(3)(2)(1)
Answered by Reiny
why did you start with n! ?

it said:
(n-r)!/(n-r-1)!

which is
(n-r)(n-r-1)(n-r-2) .. (3)(2)(1) /((n-r-1)(n-r-2)...(3)(2)(1)
= n-r
Answered by Joey
Oh ok, thank you! I kind of understand it now and will do some extra questions.
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