In mathematics, permutation connects to the process of collecting all the partners of a party into some sequence or format. In different phrases, if the party is already executed, then the redirecting of its members is called the process of permuting. Permutations take place, in more or less significant ways, in nearly every community of mathematics. They often occur when distinct management on specific limited areas is observed.
Permutation
It is the distinct interpretations of a provided number of components carried one by one, or some, or all at a time. For example, if we have two components A and B, then there are two likely performances, AB and BA.
A numeral of permutations when ‘r’ components are positioned out of a total of ‘n’ components is
nPr = n! / (n – r)!
For example, let n = 3 (A, B, and C) and r = 2 (All permutations of size 2). The answer is 3!/(3 – 2)! = 6. The six permutations are AB, AC, BA, BC, CA, and CB.
Explanation of Permutation formula
A permutation is a type of performance that indicates how to permute. If there are three different numerals 1, 2, and 3, and if someone is curious to permute the numerals taking 2 at a moment, it shows (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be accomplished in 6 methods.
Here, (1, 2) and (2, 1) are distinct. Again, if these 3 numerals shall be put handling all at a time, then the interpretations will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways.
In general, n distinct things can be set taking r (r < n) at a time in n(n – 1)(n – 2)…(n – r + 1) ways. In fact, the first thing can be any of the n things. Now, after choosing the first thing, the second thing will be any of the remaining n – 1 thing. Likewise, the third thing can be any of the remaining n – 2 things. Alike, the rth thing can be any of the remaining n – (r – 1) things.
Hence, the entire number of permutations of n distinct things carrying r at a time is n(n – 1)(n – 2)…[n – (r – 1)] which is written as n Pr. Or, in other words,
nPr = n!/(n – r)!
Combination
It is the distinct sections of a shared number of components carried one by one, or some, or all at a time. For example, if there are two components A and B, then there is only one way to select two things, select both of them.
Number of combinations when ‘r’ components are chosen out of a total of ‘n’ components is, nCr = n! / [(r!) x (n – r)! ].
For example, let n = 3 (A, B, and C) and r = 2 (All combinations of size 2). The answer is 3!/((3 – 2)! × 2!) = 3. The six combinations are AB, AC, and BC.
nCr = nC(n-r)
Note: In the same example, we have distinct points for permutation and combination. For, AB and BA are two distinct items but for selecting, AB and BA are the same.
Explanation of Combination Formula
Combination, on the further hand, is a type of pack. Again, out of those three numbers 1, 2, and 3 if sets are created with two numbers, then the combinations are (1, 2), (1, 3), and (2, 3).
Here, (1, 2) and (2, 1) are identical, unlike permutations where they are distinct. This is written as 3C2. In general, the number of combinations of n distinct things taken r at a time is,
nCr = n! /[r! × (n – r)!] = nPr/r!
How many 4 letter codewords can be made from the alphabet with all letters unique?
Solution:-
There are a total of 26 letters in the English alphabet now we have to select 4 letter code and repetition is not allowed
Now, for the first place we have 26 choices for the second place, we have 25 choices for the third place we have 24 choices, and for the fourth place we have 23 choices
= 26×25×24×23 = 26!/22! = 358,800
Similar Problems
Question 1: Find the number of permutations and combinations of n = 9 and r = 3.
Solution:
Given,
n = 9
r = 3
Using the formula given above:
Permutation:
nPr = (n!) / (n – r)!
= (9!) / (9 – 3)!
= 9! / 6! = (9 × 8 × 7 × 6! )/ 6!
= 504
Combination:
nCr = n!/r!(n − r)!
= 9!/3!(9 − 3)!
= 9!/3!(6)!
= 9 × 8 × 7 × 6!/3!(6)!
= 84
Question 2: In how many ways a committee consisting of 4 men and 2 women, can be chosen from 6 men and 5 women?
Solution:
Choose 4 men out of 6 men = 6C4 ways = 15 ways
Choose 2 women out of 5 women = 5C2 ways = 10 ways
The committee can be chosen in 6C4 × 5C2 = 150 ways.
Question 3: How considerable words can be created by using 2 letters from the term“LOVE”?
Solution:
The term “LOVE” has 4 distinct letters.
Therefore, required number of words = 4P2 = 4! / (4 – 2)!
Required number of words = 4! / 2! = 24 / 2 = 12
Question 4: Out of 5 consonants and 3 vowels, how many words of 3 consonants and 2 vowels can be formed?
Solution:
A number of ways of choosing 3 consonants from 5.
= 5C3
A number of ways of choosing 2 vowels from 3.
= 3C2
A number of ways of choosing 3 consonants from 2 and 2 vowels from 3.
= 5C3 × 3C2
= 10 × 3
= 30
It means we can have 30 groups where each group contains a total of 5 letters (3 consonants and 2 vowels).
Number of ways of arranging 5 letters among themselves
= 5! = 5 × 4 × 3 × 2 × 1 = 120
Hence, the required number of ways
= 30 × 120
= 3600
Question 5: How many different combinations do you get if you have 5 items and choose 4?
Solution:
Insert the given numbers into the combinations equation and solve. “n” is the number of items that are in the set (5 in this example); “r” is the number of items you’re choosing (4 in this example):
C(n, r) = n! / r! (n – r)!
= 5! / 4! (5 – 4)!
= (5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1 × 1)
= 120/24
= 5
Question 6: Out of 6 consonants and 3 vowels, how many expressions of 2 consonants and 1 vowel can be created?
Solution:
A number of ways of selecting 2 consonants from 6.
= 6C2
A number of ways of selecting 1 vowel from 3.
= 3C1
A number of ways of selecting 3 consonants from 7 and 2 vowels from 4.
= 6C2 × 3C1
= 15 × 3
= 45
It means we can have 45 groups where each group contains a total of 3 letters (2 consonants and 1 vowel).
A number of ways of arranging 3 letters among themselves.
= 3! = 3 × 2 × 1
= 6
Hence, the required number of ways.
= 45 × 6
= 270
Question 7: In how many distinct forms can the letters of the term ‘PHONE’ be organized so that the vowels consistently come jointly?
Solution:
The word ‘PHONE’ has 5 letters. It has the vowels ‘O’,’ E’, in it and these 2 vowels should consistently come jointly. Thus these two vowels can be grouped and viewed as a single letter. That is, PHN(OE).
Therefore we can take total letters like 4 and all these letters are distinct.
A number of methods to organize these letters.
= 4! = 4 × 3 × 2 × 1
= 24
All the 2 vowels (O, E) are distinct.
A number of ways to arrange these vowels among themselves.
= 2! = 2 × 1
= 2
Hence, the required number of ways.
= 24 × 2
= 48.
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