Question

What is the value of 7P3 ? A. 21 B. 35 C. 210 D. 840

i think a but idk

Answer 1

The answer is C.
Here's why, nPr = n!/(n-r)!. Plug in 7 for n and 3 for r. And you get 7!/(7-3)!.
Thus, 7!/4! = (7*6*5*4*3*2*1)/(4*3*2*1), which simplifies to 7*6*5 = 210

Answer 2

Answer  

Option (c) is correct

The value of  $^7P_3=210$

Step-by-step explanation:

 Given : $^7P_3$

We have to find the value of  $^7P_3$

Consider $^7P_3$

Permutation is defined as number of possibilities for choosing an ordered set  of r objects from a total of n objects.

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

put n = 7 and r = 3

We have,

$^7P_3=\frac{7!}{\left(7-3\right)!}$

Simplify, we have

$=\frac{7!}{4!}$

$\quad \frac{n!}{\left(n-m\right)!}=n\cdot \left(n-1\right)\cdots \left(n-m+1\right),\:n>m$

$\frac{7!}{4!}=7\cdot \:6\cdot \:5=210$

Thus, The value of  $^7P_3=210$