Question

Solve ,nC2 = 4n+5 algebraically.

Answer 1

n = 10

Explanation:

Using the definition of n
`C_(r)` =
`(n!)/(r!(n-r)!)`

where n! = n(n - 1)(n - 2)..... × 3 × 2 × 1

Given

n
`C_(2)` = 4n + 5 , then

`(n(n-1)(n-2))/(2!(n-2)!)` = 4n + 5

Cancel the terms from (n - 2).... on numerator/denominator

`(n(n-1))/(2)` = 4n + 5 ( multiply both sides by 2 )

n² - n = 8n + 10 ( subtract 8n + 10 from both sides )

n² - 9n - 10 = 0 ← in standard form

(n - 10)(n + 1) = 0 ← in factored form

Equate each factor to zero and solve for n

n - 10 = 0 ⇒ n = 10

n + 1 = 0 ⇒ n = - 1

However n > 0 ⇒ n = 10