Question

How do you evaluate ${}^9{P_3}$?

Answer 1

Hint: use the formula for permutation and substitute the values of n and r in the formula. After substituting the required values solve the factorial to get the desired answer.

Complete step by step solution:
In the above question we are required to evaluate the value of $^9{P_3}$
We know that,
$^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
On comparing the given equation with the above formula we come to the conclusion that
$n = 9$ and
$r = 3$
Now substitute the values of n and r in the formula and we reach:
$^9{P_3} = \dfrac{{9!}}{{(9 - 3)!}}$
On simplifying we get
$^9{P_3} = \dfrac{{9!}}{{(6)!}}$
Which can also be written as
$^9{P_3} = \dfrac{{9 \times 8 \times 7 \times 6!}}{{(6)!}}$
On further simplifying we get
$^9{P_3} = 9 \times 8 \times 7$

$^9{P_3} = 504$

Note:
Factorial is defined as the product of all natural numbers less than or equal to a given number.
For the natural number ‘n’, it’s factorial is denoted as n!
And $n! = n(n - 1)(n - 2)(n - 3)(n - 4)....3 \times 2 \times 1$
Remember that:
$
  0! = 1 \\
  1! = 1 \\
 $
And
The permutation is the arrangement of objects in a definite order.
The number of permutations of n different objects taken r at a time without replacement, where $0 < r \leqslant n$, is given by:
$^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$
The selection of some or all objects from a given set of different objects where the order of selection is not considered is called Combination. Therefore, the number of combinations of n different objects taken r objects out of them without replacement, $0 < r \leqslant n$, is given by:
$^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} = \dfrac{{^n{P_r}}}{{r!}}$
Remember:
If in a problem statement, you are asked for selection and their ordering, then you should use permutation.
In simple words, we can say that,
Permutation=Selecting + Ordering
If in any problem statement, you are asked only for selection then you should use a combination.
In simple words, we say that,
Combination= Selection.