Binomial Theorem (1 Viewer)

[cutemouse]

Hello, could someone please help me with the following question. Thanks

[maths]\text{Find } \text{the } \text{value } \text{of } n \text{ so } \text{that }:\\^nC_3-\left(2.^{n+1}C_2\right)=^nC_1[/maths]

 

[study-freak]

n!/3!(n-3)!-2(n+1)!/(n-1)!2!=n!/(n-1)!
n(n-1)(n-2)/3!-2(n+1)n/2!=n
(n-1)(n-2)/3!-2(n+1)/2!=1
n^2-3n+2-6(n+1)=6
n^2-9n-10=0
(n-10)(n+1)=0
n=10 or -1
Edit:Whoops, n can't be -1
Thus, n=10

 

[jet]