Solve the equation C(n+2,4)=n^2-1 .
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
We have to solve : C(n+2,4) = n^2-1
We know that C(n,k) = n!/k!(n-k)!
C(n+2,4) = n^2-1
=> (n +2)! / 4! * ( n - 2)! = n^2 - 1
=> n*(n -1)(n + 1)((n +2) / 4! = ( n - 1)( n + 1)
=> n(n + 2) = 4!
=> n^2 + 2n = 24
=> n^2 + 2n - 24 = 0
=> n^2 + 6n - 4n - 24 = 0
=> n( n + 6) - 4( n + 6) = 0
=> (n - 4)(n + 6) = 0
n - 4 = 0, gives n = 4
n + 6 = 0, gives n = -6
As the factorial is defined for only a positive number we take n = 4.
The required value of n = 4
giorgiana1976 | College Teacher | (Level 3) Valedictorian
We'll write the formula for combinations:
C(n,k) = n!/k!(n-k)!
According to this formula, we'll write the combinations for the given equation:
C(n+2,4) = (n+2)!/4!(n+2-4)!
C(n+2,4) = (n+2)!/4!(n-2)!
But (n+2)! = (n-2)!*(n-1)*n*(n+1)*(n+2)
We notice in this product the presence of the difference of squares:
(n-1)*(n+1) = n^2 - 1
C(n+2,4) = (n-2)!*(n^2 - 1)*n*(n+2)/4!*(n-2)!
We'll re-write the equation:
(n^2 - 1)*n*(n+2)/4! = n^2 - 1
We'll divide by n^2 - 1:
n*(n+2)/4! = 1
n(n+2) = 4!
We'll remove the brackets and we'll substitute 4!:
n^2 + 2n = 1*2*3*4
n^2 + 2n - 24 = 0
We'll apply quadratic formula:
n1 = [-2+sqrt(4+96)]/2
n1 = (-2+10)/2
n1 = 4
n2 = -12/2
n2 = -6
Since n has to be a natural number, the only solution of the equation is n = 4.