SOLUTION: hey, i would greatly appreciate it if you could solve this combinatorics problem for me
(n+2)!/n! = 56
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Question 154621: hey, i would greatly appreciate it if you could solve this combinatorics problem for me
(n+2)!/n! = 56
Found 2 solutions by edjones, jim_thompson5910:
Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
(n+2)!/n! = 56
((n+2)(n+1)*n!)/n!=56
(n+2)(n+1)=56
n^2+3n+2=56
n^2+3n-54=0
(n+9)(n-6)=0
n=-9, n=6
.
Ed
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
!}{n!}=56)
Start with the given equation.
\left(n+2-1\right)\left(n+2-2\right)\left(n+2-3\right)\cdots\left(3\right)\left(2\right)\left(1\right)}{n!}=56)
Expand . Notice how each successive term is going down by 1
\left(n+2-1\right)\left(n+2-2\right)\left(n+2-3\right)\left(n+2-4\right)\left(n+2-5\right)\left(n+2-6\right)\cdots\left(3\right)\left(2\right)\left(1\right)}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\cdots\left(3\right)\left(2\right)\left(1\right)}=56)
Expand
\left(n+1\right)\left(n\right)\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\cdots\left(3\right)\left(2\right)\left(1\right)}{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\cdots\left(3\right)\left(2\right)\left(1\right)}=56)
Subtract and simplify
Cancel out the common terms
\left(n+1\right)}{1}=56)
Simplify
Reduce
FOIL the left side
Subtract 56 from both sides.
Factor the left side
or Set each factor equal to zero
or Solve for "n"
So the possible solutions are or . However, you cannot evaluate the factorial of a negative number (well not yet anyway). So the only solution is
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