Question

If ${}^{12}{{P}_{r}}=1320$, then $r$ is equal to $?$

& \text{1) 5} \\\ & \text{2) 4} \\\ & \text{3) 3} \\\ & \text{4) 2} \\\ \end{aligned}$$

Answer 1

Hint : Using the formula of permutations of $n$ different things taken $r$ at a time and by finding the factorials of the given term then equating the equations. We can find out the value of $r$ and we can check that out of the given four options which one option is correct.

Complete step-by-step solution:
In daily life , sometimes we are required to know the number of ways which are useful to decide a particular thing in different ways .To find the number of ways of doing work we have used a counting technique by which we can select the objects.
The product of first $n$ natural numbers is called factorial $n$ (or sometimes, $n$ factorial). It is denoted by either of the symbols $n!$ or $\left| \\!{\nderline {\, n \,}} \right.$
In $n!$ , $n$ represents the number and the $!$ indicates the factorial process.
We also define $0!=1$ thus $1!=1$, $2!=2\times 1$,$3!=3\times 2\times 1$ and so on
In general,
$n!=1\cdot 2\cdot 3....n$
$n!=n\cdot (n-1)....3\cdot 2\cdot 1$
$n!=n\cdot (n-1)!$
$n!=n(n-1)(n-2)!$
Permutations means arrangements. Each of the different arrangements which can be made by taking some or all of a number of things is called permutation. In permutation, order of appearance of things is taken into account.
Permutations can be classified into the two different categories.
Where repetition is allowed , for example coins in your pocket [1,2,4,4,5]
Where repetition is not allowed, for example a lottery number [14,23,38,44]
To find the number of permutations of $n$ different things taken $r$ at a time
A permutation is the choice of $r$ things from a set of $n$ things without replacement and order does not matter.
Where $n$ and $r$are positive integers such that $1\le r\le n$ denoted by the symbol $P(nr)$ or ${}^{n}{{P}_{r}}$.
Where, ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Here ${}^{n}{{P}_{r}}$ is the permutation ,$n$ is the total number of objects and $r$ is the number of objects selected
${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}=n(n-1)(n-2).....(n-r+1)$
The last factor is $[n-(r-1)]=n-r+1$
Where $n!=1\cdot 2\cdot 3....n$
Now according to the question
${}^{12}{{P}_{r}}=1320$
Using the formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
Where $n=12$
${}^{12}{{P}_{r}}=\dfrac{12!}{\left( 12-r \right)!}$
$\dfrac{12!}{\left( 12-r \right)!}=1320$
$\dfrac{12\times 11\times 10\times 9!}{\left( 12-r \right)!}=1320$
$\dfrac{1320\times 9!}{\left( 12-r \right)!}=1320$
Here $1320$ will be cancelled out from both sides.
$\dfrac{9!}{\left( 12-r \right)!}=1$
$\left( 12-r \right)!=9!$
On comparing both sides
$\left( 12-r \right)=9$
$r=12-9$
$r=3$
Hence option$(3)$ is correct , as by calculating we have got the value of $r=3$

Note: Remember that factorial of negative integers, fractions or irrational numbers is not defined. Factorial $n$ is defined only for $n$ whole numbers. We cannot cancel, split or combine factorials because factorial is an operation that needs to precede multiplication and division.