Permutation
Consider the following problem. Mohan has 3 pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore, there are 3 × 2 = 6 pairs of a pant and a shirt.
Example Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
Solution There are as many words as there are ways of filling in 4 vacant places by the 4 letters, keeping in mind that the repetition is not allowed. The first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the required number of words is 24.
Note If the repetition of the letters was allowed, how many words can be formed?
One can easily understand that each of the 4 vacant places can be filled in succession in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256.
Example 2 Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?
Solution There will be as many signals as there are ways of filling in 2 vacant places
in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12.
Example 3 How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?
Solution There will be as many ways as there are ways of filling 2 vacant places in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.
Example 4 Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available
Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags separately and then add the respective numbers There will be as many 2 flag signals as there are ways of filling in 2 vacant places in succession by the 5 flags available. By Multiplication rule, the number of ways is 5 × 4 = 20.
Similarly, there will be as many 3 flag signals as there are ways of filling in 3 vacant places in succession by the 5 flags.
The number of ways is 5 × 4 × 3 = 60.
Continuing the same way, we find that
The number of 4 flag signals = 5 × 4 × 3 × 2 = 120
and the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120
Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320.
Solve these problems on your own. If any problem See the examples above.
1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Answers ……… 1. (i) 125, (ii) 60. 2. 108 3. 5040 4. 336 5. 8 6. 20
7.3.1 Permutations when all the objects are distinct
Theorem 1 The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by .
Theorem 2 The number of permutations of n different objects taken r at a time, where repetition is allowed, is
7.3.4 Permutations when all the objects are not distinct objects Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, O1 and O2. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, RO1O2T. Corresponding to this permutation,we have 2 ! permutations RO1O2T and RO2O1T which will be exactly the same permutation if O1 and O2 are not treated as different, i.e., if O1 and O2 are the same O at both places Therefore, the required number of permutations
= 4!/2! =3*4=12
now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times. let us treat these letters different and name them as I1, I2, T1 , T2, T3. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !.
Consider one such permutation, say, I1 NT1 SI2 T2 U E T3. Here if I1, I2 are not same and T1, T2, T3 are not same, then I1, I2 can be arranged in 2! ways and T1, T2, T3 can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation I1NT1SI2T2UET3. Hence, total number of different permutations will be
9!/ 2!*3!
Theorem 3 The number of permutations of n objects, where p objects are of the same kind and rest are all different = n!/ p!
Example 9 Find the number of permutations of the letters of the word ALLAHABAD.
Solution Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are all different.
Therefore, the required number of arrangements =
9! /4!*2! = 7560
Example 10 How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Solution Here order matters for example 1234 and 1324 are two different numbers.
Therefore, there will be as many 4 digit numbers as there are permutations of 9 different
digits taken 4 at a time.
Therefore, the required 4 digit numbers ( )
9P4 = 9!/ (9!-4!) = 3024.
Example 11 How many numbers lying between 100 and 1000 can be formed with the
digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?
Solution Every number between 100 and 1000 is a 3-digit number. We, first, have to count the permutations of 6 digits taken 3 at a time. This number would be 6P3. But, these permutations will include those also where 0 is at the 100’s place. For example, 092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from 6P3 to get the required number. To get the number of such numbers, we fix 0 at the 100’s place and rearrange the remaining 5 digits taking 2 at a time. This number is 5P2. So The required
6P3 – 5P2 = (6!/3!) – (5!/3!)
= 4 × 5 × 6 – 4 ×5 = 100
Example 14 Find the number of different 8-letter arrangements that can be made
from the letters of the word DAUGHTER so that
(i) all vowels occur together (ii) all vowels do not occur together.
Solution (i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320.
(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together. Therefore, the required number 8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6)
= 2 × 6 ! (28 – 3)
= 50 × 6 ! = 50 × 720 = 36000
Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ?
Solution Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements
= 9! / (4! 3! 2!)
=1260
Example 16 Find the number of arrangements of the letters of the word
INDEPENDENCE. In how many of these arrangements,
(i) do the words start with P
(ii) do all the vowels always occur together
(iii) do the vowels never occur together
(iv) do the words begin with I and end in P?
Solution There are 12 letters, of which N appears 3 times, E appears 4 times and D
appears 2 times and the rest are all different. Therefore
The required number of arrangements
12!/ (3! 4! 2!)
=1663200
(i) Let us fix P at the extreme left position, we, then, count the arrangements of the
remaining 11 letters. Therefore, the required number of words starting with P are
11! / (3! 2! 4!)
= 138600
(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in 8! /(3! 2! ) ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in 5!/4! ways. Therefore, by multiplication principle the required number of Arrangements
= 8! 5! / 3! 2! 4! =16800
(iii) The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.
= 1663200 – 16800 = 1646400
(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end).
We are left with 10 letters.
Hence, the required number of arrangements
=
10! / 3!2! 4!
= 12600
Ye problem solve kro agr nhi aye to answer nhi above examples phir se dkho
1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is
repeated?
2. How many 4-digit numbers are there with no digit repeated?
3. How many 3-digit even numbers can be made using the digits
1, 2, 3, 4, 6, 7, if no digit is repeated?
4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4,
5 if no digit is repeated. How many of these will be even?
5. From a committee of 8 persons, in how many ways can we choose a chairman
and a vice chairman assuming one person can not hold more than one position?
8. How many words, with or without meaning, can be formed using all the letters of
the word EQUATION, using each letter exactly once?
9. How many words, with or without meaning can be made from the letters of the
word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time, (ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
10. In how many of the distinct permutations of the letters in MISSISSIPPI do the
four I’s not come together?
11. In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S, (ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Answers………………..
1. 504 2. 4536 3. 60 4. 120, 48
5. 56 6. 9 7. (i) 3, (ii) 4 8. 40320
9. (i) 360, (ii) 720, (iii) 240 10. 33810
11. (i) 1814400, (ii) 2419200, (iii) 25401600
Consider the following problem. Mohan has 3 pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with? There are 3 ways in which a pant can be chosen, because there are 3 pants available. Similarly, a shirt can be chosen in 2 ways. For every choice of a pant, there are 2 choices of a shirt. Therefore, there are 3 × 2 = 6 pairs of a pant and a shirt.
Example Find the number of 4 letter words, with or without meaning, which can be formed out of the letters of the word ROSE, where the repetition of the letters is not allowed.
Solution There are as many words as there are ways of filling in 4 vacant places by the 4 letters, keeping in mind that the repetition is not allowed. The first place can be filled in 4 different ways by anyone of the 4 letters R,O,S,E. Following which, the second place can be filled in by anyone of the remaining 3 letters in 3 different ways, following which the third place can be filled in 2 different ways; following which, the fourth place can be filled in 1 way. Thus, the number of ways in which the 4 places can be filled, by the multiplication principle, is 4 × 3 × 2 × 1 = 24. Hence, the required number of words is 24.
Note If the repetition of the letters was allowed, how many words can be formed?
One can easily understand that each of the 4 vacant places can be filled in succession in 4 different ways. Hence, the required number of words = 4 × 4 × 4 × 4 = 256.
Example 2 Given 4 flags of different colours, how many different signals can be generated, if a signal requires the use of 2 flags one below the other?
Solution There will be as many signals as there are ways of filling in 2 vacant places
in succession by the 4 flags of different colours. The upper vacant place can be filled in 4 different ways by anyone of the 4 flags; following which, the lower vacant place can be filled in 3 different ways by anyone of the remaining 3 different flags. Hence, by the multiplication principle, the required number of signals = 4 × 3 = 12.
Example 3 How many 2 digit even numbers can be formed from the digits 1, 2, 3, 4, 5 if the digits can be repeated?
Solution There will be as many ways as there are ways of filling 2 vacant places in succession by the five given digits. Here, in this case, we start filling in unit’s place, because the options for this place are 2 and 4 only and this can be done in 2 ways; following which the ten’s place can be filled by any of the 5 digits in 5 different ways as the digits can be repeated. Therefore, by the multiplication principle, the required number of two digits even numbers is 2 × 5, i.e., 10.
Example 4 Find the number of different signals that can be generated by arranging at least 2 flags in order (one below the other) on a vertical staff, if five different flags are available
Solution A signal can consist of either 2 flags, 3 flags, 4 flags or 5 flags. Now, let us count the possible number of signals consisting of 2 flags, 3 flags, 4 flags and 5 flags separately and then add the respective numbers There will be as many 2 flag signals as there are ways of filling in 2 vacant places in succession by the 5 flags available. By Multiplication rule, the number of ways is 5 × 4 = 20.
Similarly, there will be as many 3 flag signals as there are ways of filling in 3 vacant places in succession by the 5 flags.
The number of ways is 5 × 4 × 3 = 60.
Continuing the same way, we find that
The number of 4 flag signals = 5 × 4 × 3 × 2 = 120
and the number of 5 flag signals = 5 × 4 × 3 × 2 × 1 = 120
Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320.
Solve these problems on your own. If any problem See the examples above.
1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Answers ……… 1. (i) 125, (ii) 60. 2. 108 3. 5040 4. 336 5. 8 6. 20
7.3.1 Permutations when all the objects are distinct
Theorem 1 The number of permutations of n different objects taken r at a time, where 0 < r ≤ n and the objects do not repeat is n ( n – 1) ( n – 2). . .( n – r + 1), which is denoted by .
Theorem 2 The number of permutations of n different objects taken r at a time, where repetition is allowed, is
7.3.4 Permutations when all the objects are not distinct objects Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, O1 and O2. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, RO1O2T. Corresponding to this permutation,we have 2 ! permutations RO1O2T and RO2O1T which will be exactly the same permutation if O1 and O2 are not treated as different, i.e., if O1 and O2 are the same O at both places Therefore, the required number of permutations
= 4!/2! =3*4=12
now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times. let us treat these letters different and name them as I1, I2, T1 , T2, T3. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !.
Consider one such permutation, say, I1 NT1 SI2 T2 U E T3. Here if I1, I2 are not same and T1, T2, T3 are not same, then I1, I2 can be arranged in 2! ways and T1, T2, T3 can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation I1NT1SI2T2UET3. Hence, total number of different permutations will be
9!/ 2!*3!
Theorem 3 The number of permutations of n objects, where p objects are of the same kind and rest are all different = n!/ p!
Example 9 Find the number of permutations of the letters of the word ALLAHABAD.
Solution Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are all different.
Therefore, the required number of arrangements =
9! /4!*2! = 7560
Example 10 How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?
Solution Here order matters for example 1234 and 1324 are two different numbers.
Therefore, there will be as many 4 digit numbers as there are permutations of 9 different
digits taken 4 at a time.
Therefore, the required 4 digit numbers ( )
9P4 = 9!/ (9!-4!) = 3024.
Example 11 How many numbers lying between 100 and 1000 can be formed with the
digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?
Solution Every number between 100 and 1000 is a 3-digit number. We, first, have to count the permutations of 6 digits taken 3 at a time. This number would be 6P3. But, these permutations will include those also where 0 is at the 100’s place. For example, 092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from 6P3 to get the required number. To get the number of such numbers, we fix 0 at the 100’s place and rearrange the remaining 5 digits taking 2 at a time. This number is 5P2. So The required
6P3 – 5P2 = (6!/3!) – (5!/3!)
= 4 × 5 × 6 – 4 ×5 = 100
Example 14 Find the number of different 8-letter arrangements that can be made
from the letters of the word DAUGHTER so that
(i) all vowels occur together (ii) all vowels do not occur together.
Solution (i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations = 6 ! × 3 ! = 4320.
(ii) If we have to count those permutations in which all vowels are never together, we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together. Therefore, the required number 8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6)
= 2 × 6 ! (28 – 3)
= 50 × 6 ! = 50 × 720 = 36000
Example 15 In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ?
Solution Total number of discs are 4 + 3 + 2 = 9. Out of 9 discs, 4 are of the first kind (red), 3 are of the second kind (yellow) and 2 are of the third kind (green).
Therefore, the number of arrangements
= 9! / (4! 3! 2!)
=1260
Example 16 Find the number of arrangements of the letters of the word
INDEPENDENCE. In how many of these arrangements,
(i) do the words start with P
(ii) do all the vowels always occur together
(iii) do the vowels never occur together
(iv) do the words begin with I and end in P?
Solution There are 12 letters, of which N appears 3 times, E appears 4 times and D
appears 2 times and the rest are all different. Therefore
The required number of arrangements
12!/ (3! 4! 2!)
=1663200
(i) Let us fix P at the extreme left position, we, then, count the arrangements of the
remaining 11 letters. Therefore, the required number of words starting with P are
11! / (3! 2! 4!)
= 138600
(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object EEEEI for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in 8! /(3! 2! ) ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in 5!/4! ways. Therefore, by multiplication principle the required number of Arrangements
= 8! 5! / 3! 2! 4! =16800
(iii) The required number of arrangements = the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.
= 1663200 – 16800 = 1646400
(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end).
We are left with 10 letters.
Hence, the required number of arrangements
=
10! / 3!2! 4!
= 12600
Ye problem solve kro agr nhi aye to answer nhi above examples phir se dkho
1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is
repeated?
2. How many 4-digit numbers are there with no digit repeated?
3. How many 3-digit even numbers can be made using the digits
1, 2, 3, 4, 6, 7, if no digit is repeated?
4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4,
5 if no digit is repeated. How many of these will be even?
5. From a committee of 8 persons, in how many ways can we choose a chairman
and a vice chairman assuming one person can not hold more than one position?
8. How many words, with or without meaning, can be formed using all the letters of
the word EQUATION, using each letter exactly once?
9. How many words, with or without meaning can be made from the letters of the
word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time, (ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
10. In how many of the distinct permutations of the letters in MISSISSIPPI do the
four I’s not come together?
11. In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S, (ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Answers………………..
1. 504 2. 4536 3. 60 4. 120, 48
5. 56 6. 9 7. (i) 3, (ii) 4 8. 40320
9. (i) 360, (ii) 720, (iii) 240 10. 33810
11. (i) 1814400, (ii) 2419200, (iii) 25401600
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