Question

permutation and probability problem:
its attached

Answer 1

This is not probability.

Answer 2

\[nPr = \frac{n!}{(n-r)!}\]
So...
\[90 \times _{(n-1)}P_8 = _{(n+1)}P_9 \]\[90 \times \frac{(n-1)!}{[(n-1)-8)]!} = \frac{(n+1)!}{[(n+1)-9)]!} \]\[90 \times \frac{(n-1)!}{(n-9)!} = \frac{(n+1)!}{(n-8)!} \]\[90 \times \frac{(n-1)(n-2)(n-3)...(n-9)!}{(n-9)!} = \frac{(n+1)(n)(n-1)...(n-8)!}{(n-8)!} \]\[90 \times (n-1)(n-2)(n-3)...(n-8) = (n+1)(n)(n-1)...(n-7) \]\[90 (n-8) = (n+1)(n)\]\[90 n-720 = n^2+n\]\[0= n^2-89n+720\]\[n = 9 \ or \ n = 80\]

Answer 3

Don't know if it works :S

Answer 4

those are correct