Question

find n if , factorial n / factorial n-2 =110​

Answer 1

We have to find n if,

nP2 = 110

We know,

nPr = (n!)/((n - r)!)

And also, we remember that n and r are positive integers.

So,

nP2 = 110

=> (n!)/((n - 2)!) = 110

=> (1 • 2 • 3 • ... • n)/(1 • 2 • 3 • ... • (n - 2)) = 110

=> (1 • 2 • 3 • ... • (n - 2)(n - 1)n)/(1 • 2 • 3 • ... • (n - 2)) = 110

=> n(n - 1) = 110

=> n^2 - n = 110

=> n^2 - n - 110 = 0

=> n^2 - 11n + 10n - 110 = 0

=> n(n - 11) + 10(n - 11) = 0

=> (n + 10)(n - 11) = 0

=> n = -10 ; n = 11

But since n is a positive integer,

n = 11

Hence 11 is the answer.

Answer 2

Answer: n = 11

Step-by-step explanation:

Given,

$^nP_2=110\\\;\\\frac{n!}{(n-2)!}=110\\\;\\\frac{n(n-1)(n-2)!}{(n-2)!}=110\\\;\\n(n-1)=110\\\;\\n^2-n=110\\\;\\n^2-n-110=0\\\;\\n^2+10n-11n-110=0\\\;\\n(n+10)-11(n+10)=0\\\;\\(n-11)(n+10)=0\\\;\\We\;have,\\\;\\n=11\;\;\;or\;\;\;n=-10$

But We know that, n can not we negative in case of permutation and combination.

Hence, n = 11

Note:-

$1.\;\;^nP_r=\frac{n!}{(n-r)!}\\\;\\2.\;\;n!=n(n-1)(n-2)..........3\times2\times1$