What is the value of n?

If 2nC4: nC3 = 21:1, then find the value of n.
please note that this is a permutation combination question.

1 Answer
Apr 23, 2018

n=5

Explanation:

2nC4=((2n)!)/(4!(2n-4)!)

nC3=(n!)/(3!(n-3)!)

(2nC4)/(nC3)=(((2n)!)/(4!(2n-4)!))/((n!)/(3!(n-3)!))

=((2n)!3!(n-3)!)/(4!n!(2n-4)!)

Notice a few things.

(3!)/(4!)=1/4,

((2n)!)/((2n-4)!)=2n(2n-1)(2n-2)(2n-3),

and

((n-3)!)/(n!)=1/(n(n-1)(n-2)).

Putting this all together we have

((2n)!3!(n-3)!)/(4!n!(2n-4)!)=(2n(2n-1)(2n-2)(2n-3))/(4n(n-1)(n-2))

=(4n(2n-1)(n-1)(2n-3))/(4n(n-1)(n-2))

=((2n-1)(2n-3))/(n-2).

We want

((2n-1)(2n-3))/(n-2)=21

(2n-1)(2n-3)=21(n-2)

4n^2-8n+3=21n-42

4n^2-29n+45=0

Quadratic formula tells us that

n=(29pmsqrt(29^2-4(4)(45)))/(2(4))=(29pm11)/8

So n=5 or n=9/4.

Since n must be an integer, n must be 5.