Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# How many ways can you arrange the letters in the word "FACTOR"?

Last updated date: 15th Jul 2024
Total views: 380.4k
Views today: 9.80k
Verified
380.4k+ views
Hint: Here, we are required to arrange the letters in the given word ‘FACTOR’. Thus, we will use Permutations to ‘arrange’ the letters keeping in mind that all the letters in the given word are unique. Thus, applying the formula and solving the factorial, we will be able to find the required ways of arrangement of letters of the given word.

Formula Used:
We will use the following formulas:
1. ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ , where $n$ is the total number of letters and $r$ represents the number of letters to be arranged.
2. $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$.

In order to find the arrangement of the word ‘FACTOR’,
First of all, we will observe that all the letters in this given word are unique and no word is the same or duplicate. Also, the number of letters in the word ‘FACTOR’ is 6.
Therefore, we will use Permutations to ‘arrange’ the 6 letters of the given word.
Thus, the formula is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
Where, $n$ is the total number of letters and $r$ represents the number of letters to be arranged, i.e. $6$ in each case.
Thus, we get,
${}^6{P_6} = \dfrac{{6!}}{{\left( {6 - 6} \right)!}} = \dfrac{{6!}}{{0!}} = 6!$
Because, $0! = 1$
Now, the formula of expanding factorial is $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$.
Hence, we get,
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 30 \times 24 = 720$

Therefore, we can arrange the letters in the word ‘FACTOR’ in 720 ways.
Thus, this is the required answer.

Note:
While solving this question, we should know the difference between permutations and combinations. Permutation is an act of arranging the numbers whereas combination is a method of selecting a group of numbers or elements in any order. Hence, Permutations and Combinations play a vital role to solve these types of questions. . Also, in order to answer this question, we should know that when we open a factorial then, we write it in the form of: $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$ as by factorial we mean that it a product of all the positive integers which are less than or equal to the given number but not less than 1.