solve for n
a. c(n,2)=28
b. p(n,5)=20p(n,3)
I answered several question of a similar type last night.
You just have to use the definition of C(n,r) and P(n,r)
for the last one:
n!/(n-5)! = 20n!/(n-3)!
n(n-1)(n-2)(n-3)(n-4) = 20n(n-1)(n-2)
divide by n(n-1)(n-2)
(n-3)(n-4) = 20
n^2 - 7n -8 = 0
(n-8)(n+1) = 0
n=8 or n=-1
are both answers valid ???
What do you think?