Solve the equation C(n+2,4)=n^2-1 .

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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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's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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