Question

If $\rm ^{2}P_{3} : \rm ^{n}P_{6} = 1 : 210$, find n.

Answer 1

your question is incorrect. A correct question is $\rm ^{n}P_{3} : \rm ^{n}P_{6} = 1 : 210$

we know, $\rm^{n}P_{r}=\frac{n!}{(n-r)!}$

so, $\rm^{n}P_{3}=\frac{n!}{(n-3)!}$

and $\rm^{n}P_{6}=\frac{n!}{(n-6)!}$

so, n!/(n - 3)! : n!/(n - 6)! = 1 : 210

or, (n - 6)!/(n - 3)! = 1/210

or, (n - 6)!/(n - 3)(n - 4)(n - 5)(n - 6)! = 1/210

or, 1/(n - 3)(n - 4)(n - 5) = 1/210

prime factors of 210 = 2 × 3 × 5 × 7

we can write 210 = 5 × 6 × 7

= (10 - 5) × (10 - 4) × (10 - 3)

so, 210 = (10 - 3) × (10 - 4) × (10 - 5)

now, 1/(n - 3)(n - 4)(n - 5) = 1/(10 - 3)(10 - 4)(10 - 5)

hence, n = 10