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

1 Answer
May 27, 2018

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