Question

if npr =1320 and ncr=220 then find n

Answer 1

n!/(n-r)!=1320,n!/r!(n-r)!=220
on dividing ,r!=6
3×2×1=r!=3!
r=3
n!/(n-r)!=1320
n!/(n-3)!=1320
n(n-1)(n-2)(n-3)!/(n-3)!=1320
n(n-1)(n-2)=1320
12×11×10=1320
so n=12

Answer 2

If npr =1320 and ncr=220 then value of "n" is 12

Solution:

The number of permutations of n objects taken r at a time is given as:

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

The number of combinations of r objects taken from a set of n objects is given as :

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

Given that npr = 1320 and ncr = 220

$\begin{array}{l}{\text { So } \frac{n P r}{n C r}=\frac{1320}{220}} \\\\ {\frac{\frac{n !}{(n-r) !}}{\frac{n !}{(n-r) ! r !}}=\frac{1320}{220}}\end{array}$

On dividing we get,

r! = 6

r! = 3!

r = 3

So put the value of r in npr formula

npr = 1320

$\frac{n !}{(n-3) !}=1320$

$n \times(n-1) \times(n-2)=12 \times 11 \times 10$

So n = 12

Hence the value of "n" is 12

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