Find the number of ways in which the letter of the word palputta can be arranged

• Forum
• Algebra
• In how many ways can the letters of the word...

Vikash Kumar Singh,

5 years ago

Grade:12th pass

FOLLOW QUESTION

We will notify on your mail & mobile when someone answers this question.

Enter email id Enter mobile number

1 Answers

25758 Points

5 years ago

The word PATALIPUTRA has total 11 characters. Out of them, 
5 are Vowels (A A I U A) occupying positions 2,4,6,8,11
6 are consonants (P T L P T R) occupying positions 1,3,4,5,8,9

 Relative order of vowels and consonants cannot be changed.
 i.e., position of vowels and consonants cannot be changed

5 vowels can be arranged in 5! / 3! ways  (we divide by 3! as there are 3 'A')

6 consonants can be arranged in 6! / 2!×2! ways (we divide by (2! × 2!) as there are 2 'P' and 2 'T')

Total number of ways = (5!/ 3!)×(6!/2!×2!)=20×180=3600

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

View courses by askIITians

Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !

Select Grade Select Subject for trial

BOOK A FREE TRIAL

Dear ,

Your Answer has been Successfully Posted!

Answer

Verified

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. $n! = n \times \left( {n - 1} \right) \times \left( {n - 2} \right) \times ...... \times 3 \times 2 \times 1$.

Complete step-by-step answer:
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.

How many ways can the letters of the word patliputra be arranged?

What is the number of ways in which the letters of the word able?

How many ways can the letters of the word MACHINE be arranged?

How many ways can the letter of the word language we are arranged in such a way that the vowels always come together?