Question

If \[{}^{12}{{P}_{r}}=1320\], then \[r\] is equal to \[?\]
\[\begin{align}
  & \text{1) 5} \\
 & \text{2) 4} \\
 & \text{3) 3} \\
 & \text{4) 2} \\
\end{align}\]

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.