Find n so that the number of nP2 is 6 times as great as the number of permutations of n-5 elements taken 2 at a time.

Question

amistre64 · Answer

maybe ....

n(n-1) = 6(n-5)(n-6)

amistre64 · Answer

got me 6 on the worng side i think

anonymous · Answer

$$_{n} P_{2} = 6\left( _{n-5}P_{2} \right)$$ $$\frac{ n! }{ \left( n-2 \right)! }=6 \left( \frac {(n-5)!} {[(n-5)-2]!} \right)$$ $$\frac{ n! }{ \left( n-2 \right)! }=6 \left( \frac {(n-5)!} {(n-7)!} \right)$$ $$n (n-1)=6 (n-5)(n-6)$$ $$n^2-n=6 (n^2-11n+30)$$ $$n^2-n=6n^2-66n+180$$ $$0=5n^2-65n+180$$ $$0=5(n^2-13n+36)$$ $$0=5(n-4)(n-9)$$ $$n = 4$$ or  $$n=9$$.

Since $$n \neq 4$$ because $$ n \ge 7$$ then $$ n= 9$$

amistre64 · Answer

yeah, the 6 on the right is gooder

anonymous · Answer

Check. 
$$ _{9}P_{7} = 6(_{4}P_{2})$$ $$9(8) = 6(4)(3) $$ $$72 = 72 $$