Question

2nC3:nC2=44:3? Permutation And book answer is 6

Answer 1

`n=6`

Explanation:

We seek `n in NN` st `""_(2n)C^3:_nC^2=44:3`

Using the definition of `""_nC^r`

`""_nC^r = (n!)/(r!(n-r)!)`

We can write the given ratio expression as :

`((2n)!)/(3!(2n-3)!) \ : \ (n!)/(2!(n-2)!) = 44 \ : \ 3`

Or, Equivalently as a fraction:

`( ((2n)!)/(3!(2n-3)!) ) / ( (n!)/(2!(n-2)!) )= 44 / 3`

So that:

`((2n)!)/(3!(2n-3)!) * (2!(n-2)!)/(n!) = 44 / 3`

And if we expand:

`((2n)(2n-1)(2n-2)(2n-3)....1)/(3!(2n-3)!) * (2!(n-2)!)/(n(n-1)(n-2)...1) = 44 / 3`

Cancelling terms, we get:

`((2n)(2n-1)(2n-2))/(3!) * (2!)/(n(n-1)) = 44 / 3`

And Simplifying:

`(2(n)(2n-1)2(n-1))/(6) * (2)/(n(n-1)) = 44 / 3`

`:. (4(2n-1))/(3) = 44 / 3`

`:. 2n-1 = 11`

`:. n=6`

And we can quickly validate the solutions:

`""_(12)C^3 \ : \ _6C^2 = 220:15 = 44:3`