Question

10pr=5040 find out r

Answer 1

n=10 r=? 
npr= n!/(n-r)!
5040= 10!/(10-r)!
7! = 10!/(10-r)!
(10-r)!=  10!/7!
(10-r)!= 10*9*8*7!/7!
(10-r)!=10*9*8
(10-r)!=720
(10-r)!=6!
10-r=6 
r=10-6
r=4

Answer 2

$^{10} \mathrm{P_r} = 5040$ then the value of r is 4

Solution:

Given that $^{10} \mathrm{P_r} = 5040$

To find: the value of r

The formula for permutation is given as:

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

Now substitute n = 10 and $^{10} \mathrm{P_r} = 5040$

$10 \mathrm{P}_{\mathrm{r}}=\frac{10 !}{(10-r) !}=5040$

On rearranging we get,

$\frac{1}{(10-r) !}=\frac{5040}{10 !}$

$\frac{1}{(10-r) !}=\frac{5040}{10 \times 9 !}$

$=\frac{504}{9 \times 8 !}=\frac{504}{9 \times 8 !}$

$=\frac{56}{8 \times 7 !}=\frac{7}{7 !}=\frac{1}{6 !}$

On comparing we get:

$\begin{array}{l}{(10-r) !=6 !} \\\\ {10-r=6} \\\\ {10-6=r} \\\\ {4=r}\end{array}$

Hence, the value of r is 4

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