Sample Problems on Permutation Formula
Question 1: What are the types of permutations?
Solution:
The permutation of a collection of things or components in order relies on three conditions:
- When recurrence of essences is not allowed
- When recurrence of essences is allowed
- When the components of a group are not different
Question 2: Calculate the number of permutations of n = 5 and r = 2.
Solution:
Given,
n = 5
r = 2
Using the formula given above:
Permutation: nPr = (n!) / (n – r)!
= (5!) / (5 – 2)!
= 5! / 3! = (5 × 4 × 3! )/ 3!
= 20
Question 3: How many 3 letter phrases with or without purpose can be created out of the letters of the word POEM when repetition of letters is not permitted?
Solution:
Here n = 4, as the word POEM has 4 letters. Since we have to create 3 letter words with or without meaning and without repetition, therefore total permutations possible are:
⇒ P(n, r) = 4!/(4 − 3)!
= 4 × 3 × 2 × 1/1
= 24
Question 4: How many 4 letter phrases with or without purpose can be created out of the letters of the word KANHA when repetition of words is permitted?
Solution:
The number of letters, in this case, is 5, as the word KANHA has 5 alphabets.
And r = 4, as a 4-letter term has to be selected.
Thus, the permutation will be:
Permutation (when repetition is permitted) = 54= 625
Question 5: It is required to place 4 men and 3 women in a row so that the women entertain the even positions. How many such configurations are feasible?
Solution:
We are given that there are 4 men and 3 women.
i.e. there are 7 positions.
The even positions are: 2nd, 4th, and the 6th places
These three places can be occupied by 3 women in P(3, 3) ways = 3!
= 3 × 2 × 1
= 6 ways
The remaining 4 positions can be occupied by 4 men in P(4, 4) = 4!
= 4 × 3 × 2 × 1
= 24 ways
Therefore, by the Fundamental Counting Principle,
Total number of ways of seating arrangements = 24 × 6
= 144
Question 6: Find the number of phrases, with or without meaning, that can be comprised with the letters of the word ‘TABLE’.
Solution:
‘TABLE’ contains 5 letters.
Thus, the numeral of phrases that can be formed with these 5 letters = 5! = 5 × 4 × 3 × 2 × 1 = 120.
Question 7: Find the number of permutations of the letters of the phrase subject such that the vowels consistently appear in odd positions.
Solution:
The word ‘SUBJECT’ has 7 letters.
There are 6 consonants and 1 vowels in it.
No. of ways 1 vowels can occur in 7 different places = 7P1 = 7 ways.
After 1 vowels take 1 place, no. of ways 6 consonants can take 6 places = 6P6 = 6! = 720 ways.
Therefore, total number of permutations possible = 720 × 720 = 518,400 ways.
Articles related to Permutation Formula:
Permutation Formula
Permutation Formula: In mathematics, permutation relates to the method of organizing all the members of a group into some series or design. In further terms, if the group is already completed, then the redirecting of its components is called the method of permuting. Permutations take place, in better or slightly effective methods, in almost every district of mathematics. They usually occur when different directions on detailed restricted sites are monitored.
Table of Content
- What is the Permutation Formula?
- Permutation Formula Explanation
- Sample Problems on Permutation Formula
- Practice Problems on Permutation Formula
- Summary – Permutation Formula