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**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.

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