Permutation and combination

Permutation and combination

 
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Fundamental Principles of Counting :

Multiplication Theorem
If an operation can be performed in m different ways and following which a second operation can be performed in n different ways, then the two operations in succession can be performed in m × n different ways.
 
Fundamental Principles of Counting :
Addition Theorem
If an operation can be performed in m different ways and a second independent operation can be performed in n different ways, either of the two operations can be performed in (m+n) ways.
 
Factorial:
Let n be a positive integer. Then n factorial (n!) can be defined as n! = n(n-1)(n-2)…1
Examples:
5! = 5 x 4 x 3 x 2 x 1 = 120 3! = 3 x 2 x 1 = 6
 
Special Cases: a)0! = 1 b)1! = 1
 
Permutations:
Permutations are the different arrangements of a given number of things by taking some or all at a time
 
Examples:
a)All permutations (or arrangements) formed with the letters a, b, c by taking three at a time are (abc, acb, bac, bca, cab, cba) b)All permutations (or arrangements) formed with the letters a, b, c by taking two at a time are (ab, ac, ba, bc, ca, cb)
 
Combinations:
Each of the different groups or selections formed by taking some or all of a number of objects is called a combination
 
Examples:
Suppose we want to select two out of three girls P, Q, R. Then, possible combinations are PQ, QR and RP. (Note that PQ and QP represent the same selection) Suppose we want to select three out of three girls P, Q, R. Then, only possible combination is PQR
 
Difference between Permutations and Combinations and How to Address a Problem
Sometimes, it will be clearly stated in the problem itself whether permutation or combination is to be used. However if it is not mentioned in the problem, we have to find out whether the question is related to permutation or combination. Consider a situation where we need to find out the total number of possible samples of two objects which can be taken from three objects P,Q , R. To understand if the question is related to permutation or combination, we need to find out if the order is important or not. If order is important, PQ will be different from QP , PR will be different from RP and QR will be different from RQ If order is not important, PQ will be same as QP, PR will be same as RP and QR will be same as RQ Hence, If the order is important, problem will be related to permutations. If the order is not important, problem will be related to combinations. For permutations, the problems can be like “What is the number of permutations the can be made”, “What is the number of arrangements that can be made”, “What are the different number of ways in which something can be arranged”, etc For combinations, the problems can be like “What is the number of combinations the can be made”, “What is the number of selections the can be made”, “What are the different number of ways in which something can be selected”, etc. Mostly problems related to word formation, number formation etc will be related to permutations. Similarly most problems related to selection of persons, formation of geometrical figures , distribution of items (there are exceptions for this) etc will be related to combinations.
 
Repetition:
The term repetition is very important in permutations and combinations. Consider the same situation described above where we need to find out the total number of possible samples of two objects which can be taken from three objects P,Q , R. If repetition is allowed, the same object can be taken more than once to make a sample. i.e., if repetition is allowed, PP, QQ, RR can also be considered as possible samples. If repetition is not allowed, then PP, QQ, RR cannot be considered as possible samples Normally repetition is not allowed unless mentioned specifically. pq and qp are two different permutations ,but they represent the same combination.
 
Number of permutations of n distinct things taking r at a time:
Number of permutations of n distinct things taking r at a time can be given by nPr = n!(n−r)!=n(n−1)(n−2)…(n−r+1)where 0≤r≤n If r > n, nPr = 0  

Special Case:
 nP0 = 1nPr is also denoted by P(n,r). nPr has importance outside combinatorics as well where it is known as the falling factorial and denoted by (n)r or nr
Examples
8P2 = 8 x 7 = 56 5P4= 5 x 4 x 3 x 2 = 120
 
Number of permutations of n distinct things taking all at a time:
Number of permutations of n distinct things taking them all at a time = nPn = n! Number of Combinations of n distinct things taking r at a time Number of combinations of n distinct things taking r at a time ( nCr) can be given by nCr = n!(r!)(n−r)!=n(n−1)(n−2)⋯(n−r+1)r!where 0≤r≤n If r > n, nCr = 0 Special Case: nC0 = 1 nCr is also denoted by C(n,r). nCr occurs in many other mathematical contexts as well where it is known as binomial coefficient and denoted by (nr)
 
Examples:
 
1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
 
ans) Number of ways of selecting 3 consonants out of 7 = 7C3 Number of ways of selecting 2 vowels out of 4 = 4C2Number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 = 7C3 x 4C2 It means that we can have 210 groups where each group contains total 5 letters(3 consonants and 2 vowels). Number of ways of arranging 5 letters among themselves = 5! = 5 x 4 x 3 x 2 x 1 = 120 Hence, Required number of ways = 210 x 120 = 25200
 
2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
 
ans) In a group of 6 boys and 4 girls, four children are to be selected such that at least one boy should be there. Hence we have 4 choices as given below We can select 4 boys ——(Option 1). Number of ways to this = 6C4 We can select 3 boys and 1 girl ——(Option 2) Number of ways to this = 6C3 x 4C1 We can select 2 boys and 2 girls ——(Option 3) Number of ways to this = 6C2 x 4C2 We can select 1 boy and 3 girls ——(Option 4) Number of ways to this = 6C1 x 4C3 Total number of ways = (6C4) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C3) = (6C2) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C1) [Applied the formula nCr = nC(n – r) ] = 15 + 80 + 90 + 24 = 209
 
3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
 
ans) From a group of 7 men and 6 women, five persons are to be selected with at least 3 men. Hence we have the following 3 choices We can select 5 men ——(Option 1) Number of ways to do this = 7C5 We can select 4 men and 1 woman ——(Option 2) Number of ways to do this = 7C4 x 6C1 We can select 3 men and 2 women ——(Option 3) Number of ways to do this = 7C3 x 6C2 Total number of ways = 7C5 + [7C4 x 6C1] + [7C3 x 6C2] = 7C2 + [7C3 x 6C1] + [7C3 x 6C2] [Applied the formula nCr = nC(n – r) ] = 21 + 210 + 525 = 756
 
4. In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?
 
ans) The word ‘OPTICAL’ has 7 letters. It has the vowels ‘O’,’I’,’A’ in it and these 3 vowels should always come together. Hence these three vowels can be grouped and considered as a single letter. That is, PTCL(OIA). Hence we can assume total letters as 5. and all these letters are different. Number of ways to arrange these letters = 5! = [5 x 4 x 3 x 2 x 1] = 120 All The 3 vowels (OIA) are different Number of ways to arrange these vowels among themselves = 3! = [3 x 2 x 1] = 6 Hence, required number of ways = 120 x 6 = 720
 
5. In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
 
ans)The word ‘CORPORATION’ has 11 letters. It has the vowels ‘O’,’O’,’A’,’I’,’O’ in it and these 5 vowels should always come together. Hence these 5 vowels can be grouped and considered as a single letter. that is, CRPRTN(OOAIO). Hence we can assume total letters as 7. But in these 7 letters, ‘R’ occurs 2 times and rest of the letters are different. Number of ways to arrange these letters = [Loading Maths… ]2520In the 5 vowels (OOAIO), ‘O’ occurs 3 and rest of the vowels are different. Hence, required number of ways = 2520 x 20 = 50400
 
6. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
 
ans)We need to select 5 men from 7 men and 2 women from 3 women Number of ways to do this = 7C5 x 3C2 = 7C2 x 3C1 [Applied the formula nCr = nC(n – r) ] = 21 x 3 = 63
 
7. In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged such that the vowels must always come together?
 
ans) The word ‘MATHEMATICS’ has 11 letters. It has the vowels ‘A’,’E’,’A’,’I’ in it and these 4 vowels must always come together. Hence these 4 vowels can be grouped and considered as a single letter. That is, MTHMTCS(AEAI). Hence we can assume total letters as 8. But in these 8 letters, ‘M’ occurs 2 times, ‘T’ occurs 2 times but rest of the letters are different. Hence,number of ways to arrange these letters = [Loading Maths… ]In the 4 vowels (AEAI), ‘A’ occurs 2 times and rest of the vowels are different. Hence, required number of ways = 10080 x 12 = 120960
 
8. There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed?
 
ans)We need to select 5 men from 8 men and 6 women from 10 women Number of ways to do this = 8C5 x 10C6 = 8C3 x 10C4 [Applied the formula nCr = nC(n – r) ] = 56 x 210 = 11760
 
9. How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
 
ans)The word ‘LOGARITHMS’ has 10 different letters. Hence, the number of 3-letter words(with or without meaning) formed by using these letters = 10P3 = 10 x 9 x 8 = 720
 
10. In how many different ways can the letters of the word ‘LEADING’ be arranged such that the vowels should always come together?
 
ans) The word ‘LEADING’ has 7 letters. It has the vowels ‘E’,’A’,’I’ in it and these 3 vowels should always come together. Hence these 3 vowels can be grouped and considered as a single letter. that is, LDNG(EAI). Hence we can assume total letters as 5 and all these letters are different. Number of ways to arrange these letters = 5! = 5 x 4 x 3 x 2 x 1 = 120 In the 3 vowels (EAI), all the vowels are different. Number of ways to arrange these vowels among themselves = 3! = 3 x 2 x 1= 6 Hence, required number of ways = 120 x 6= 720  
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