Question

The letters of the word “Flower” are taken \[4\] at a time and arranged in all possible ways. The number of arrangements that begin with ‘F’ and end with ‘R’ is
A. $ 20 $
B. $ 18 $
C. $ 12 $
D. $ 14 $

Answer 1

Hint: The letters in the word “Flower” are $ 6 $ . Out of six letters, \[4\] are taken at a time.
It is said that two letters are fixed at the first and fourth positions. $ \therefore $ two spaces are occupied.
Now, we can select any two letters out of the remaining four letters ‘ l’,’ o’, ’w’, and ‘e’.
Use “combination” to find the number of ways in which we can select two letters out of four letters.
These can be done in $ _{}^4{C_2} $ .
The formula for finding the number of combinations is $ _{}^n{C_r} $ , where $ n $ different objects taken $ r $ at a time.
 $ _{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
Here, we use factorial,
 $ \Rightarrow n! = 1 \times 2 \times 3 \times \cdots \times n $
The selected two letters are further arranged in two ways, For example, we can arrange ‘A’ and ‘B’ as AB and BA.
The total number of arrangements that begin with ‘F’ and end with ‘R’ can be found by multiplication of $ _{}^4{C_2} $ and \[2\] .

Complete step-by-step answer:
The total number of letters in the word “Flower” is $ 6 $ . We have to arrange \[4\] letters from “Flower”.
In the \[4\] letters long word, the two letters are fixed, that is the word first starts with the letter ‘F’ and ends with ‘R’.
 $ \therefore $ Two spaces are occupied.
We can select any two letters out of the remaining four letters ‘ l’,’ o’, ’w’, and ‘e’.
Apply the Combination formula;
 $ _{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
The number of combinations is $ _{}^n{C_r} $ , where $ n $ different objects taken $ r $ at a time.
We find the number of ways of selecting any two letters out of remaining four letters that is $ _{}^4{C_2} $ ,
Substitute $ n = 4 $ and $ r = 2 $ into $ _{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ .
 \[_{}^4{C_2} = \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}\]
 \[ \Rightarrow _{}^4{C_2} = \dfrac{{4!}}{{2!2!}}\]
Determine factorial to \[2!\] and \[4!\] by formula,
 $ \Rightarrow n! = 1 \times 2 \times 3 \times \cdots \times n $
To find \[2!\] , substitute $ n = 2 $ .
 $ \Rightarrow 2! = 1 \times 2 $
 $ \Rightarrow 2! = 2 $
Similarly, \[4!\] is the product of terms $ 1 $ , $ 2 $ , $ 3 $ and $ 4 $ .
 $ \Rightarrow 4! = 1 \times 2 \times 3 \times 4 $
 $ \Rightarrow 4! = 24 $
Substitute $ 2! = 2 $ and $ 4! = 24 $ into \[_{}^4{C_2} = \dfrac{{4!}}{{2!2!}}\] .
 \[_{}^4{C_2} = \dfrac{{24}}{{2 \times 2}}\]
 \[_{}^4{C_2} = 6\]
The selected two letters further arranged in two ways,
 $ \therefore $ Total number of ways to arrange \[4\] letters= $ 6 \times 2 $
 $ \Rightarrow $ Total number of ways to arrange \[4\] letters= $ 12 $

Correct Answer: C. $ 12 $

Note:
There is a common confusion among the students is the use combination and permutation. So, whenever selection is asked always use combination and for arrangement of objects always use permutation.
Formula of combination, $ _{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $
The number of combinations is $ _{}^n{C_r} $ , where $ n $ different objects taken $ r $ at a time.
Next, the factorial is the multiplication of consecutive terms from one to $ n $ . Factorial $ n! $ , gives the number of ways in which $ n $ objects can be permuted
 $ \Rightarrow n! = 1 \times 2 \times 3 \times \cdots \times n $
A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.
 $ _{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}},0 \leqslant r \leqslant n $