Find the value of n

Question

2[n-1 5]=[n-1 6]+[n-1 4]

Answer 1

I read this as 2(n-1)C5=(n-1)C6+(n-1)C4, where nCr=n!/((n-r)!r!).

For convenience let m=n-1 then we can expand:

2m!/((m-5)!5!)=m!/((m-6)!6!)+m!/((m-4)!4!).

But 6!=6*5*4!, 5!=5*4! and (m-4)!=(m-4)(m-5)*(m-6)! and (m-5)!=(m-5)*(m-6)!

So all the m! numerators cancel out.

And all the 4! in the denominators cancel out, and so do all the (m-6)!. So we're left with:

2/(5(m-5))=1/30+1/((m-4)(m-5)) which can be expressed:

12(m-4)=(m-4)(m-5)+30.

So 12m-48=m^2-9m+20+30=m^2-9m+50.

This becomes the quadratic: m^2-21m+98=(m-7)(m-14)=0.

So n=m+1=8 or 15.