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
Mr. Mike
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.

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