please help (1 Viewer)

[allybee]

im having some trouble with this question. They’re under the topic of Binomial theorem, combinations and permitations. Any help would be great

1)Find the value of n if <SUP>n+1 </SUP>P<SUB>2 </SUB>= 4n +10

 

[pLuvia]

1. ⁿ⁺¹P₂
=(n+1)!/(n-1)!
=n(n+1)(n-1)!/(n-1)!
=4n+10

So now you have
n(n+1)=4n+10
n²-3n-10=0
(n-5)(n+2)=0
n=-2,5 but n>0
Hence n=5

 

[allybee]

1. ⁿ⁺¹P₂
=(n+1)!/(n-1)!
=n(n+1)(n-1)!/(n-1)!
=4n+10

So now you have
n(n+1)=4n+10
n²-3n-10=0
(n-5)(n+2)=0
n=-2,5 but n>0
Hence n=5

thanks for your reply but how did you get from
(n+1)!/(n-1)!
to
n(n+1)(n-1)!/(n-1)!

 

[pLuvia]

Basically I expanded the (n+1)! just for the first three terms which is (n+1)(n)(n-1)!