Question

find n if n-1p3 :np4=1:9​

Answer 1

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Turn the ":" into a divide:

(n-1)P3 / nP4 = 1 / 9
=> 9 ((n-1)P3) = nP4

Then using the definition of permutation:
9 (n-1)! / (n-1-3)! = n! / (n-4)!
=> 9 (n-1)! / (n-4)! = n! / (n-4)!
but now the (n-4)! in the denominators can cancel:

=> 9 (n-1)! = n!

You should be able to do the last step yourself.

Answer 2

Answer: n = 9

Step-by-step explanation:

We know  that,

$^nP_r=\frac{(n)!}{(n-r)!}$

We have,

$^{n-1}P_3=\frac{(n-1)!}{(n-1-3)!}\\\;\\^{n-1}P_3=\frac{(n-1)!}{(n-4)!}\\\;\\^{n-1}P_3=\frac{(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}\\\;\\^{n-1}P_3=(n-1)(n-2)(n-3)$

Similarly,

$^nP_4=\frac{n!}{(n-4)!}\\\;\\^nP_4=\frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}\\\;\\^nP_4=n(n-1)(n-2)(n-3)$

Now Given that,

$\frac{^{n-1}P_3}{^nP_4}=\frac{1}{9}\\\;\\\frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac{1}{9}\\\;\\\frac{1}{n}=\frac{1}{9}\\\;\\n=9$

Hence,

n = 9