A permutation is an arrangement in a definite order of a number of objects taken some or all at a time.
The number of permutations of \(n\) different things taken \(r\) at a time (no repetition, order matters) is denoted as \({ }^n P_r\) or \(P(n, r)\) and is defined as
\({ }^n P_r=\frac{n !}{(n-r) !}, 0 \leq r \leq n\)Example 1: 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 \(={ }^9 P_4=\frac{9 !}{(9-4) !}=\frac{9 !}{5 !}=\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 !}=3024\)
The number of permutations of \(n\) different objects taken \(r\) at a time, where repetition is allowed, is \(n^r\).
Example 2: How many 3-letter words with or without meaning can be formed out of the letters of the word SMOKE when repetition of words is allowed?
Solution:
The number of objects, in this case, is 5, as the word SMOKE has 5 alphabets. and \(r=3\), as 3-letter word has to be chosen. Thus, the permutation will be:
Permutation (when repetition is allowed) \(=5^3=125\)
The number of permutations of \(n\) objects, where \(p\) objects are of the same kind and the rest are all different \(=\frac{n !}{p !}\). In fact, we have a more general formula which we can explain as:
The number of permutations of \(n\) objects, where \(p_1\) objects are of one kind, \(p_2\) are of the second kind, \(\ldots, p_k\) are of \(k^{\text {th }}\) kind and the rest, if any, are of different kind is \(\frac{n !}{p_{1} ! p_{2} ! \ldots p_{k} !}\)
Example 3: 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, and 2 L’s, and the rest are all different.
Therefore, the required number of arrangements \(=\frac{9 !}{4 ! 2 !}=\frac{5 \times 6 \times 7 \times 8 \times 9}{2}=7560\)
Example 4: 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 \({ }^6 \mathrm{P}_3\). But, these permutations will include those also where 0 is at the 100’s place. For example, \(092,042, \ldots\), etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from \({ }^6 \mathrm{P}_3\) 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 \({ }^5 \mathrm{P}_2\). So the required number
\(
\begin{aligned}
&={ }^6 \mathrm{P}_3-{ }^5 \mathrm{P}_2=\frac{6 !}{3 !}-\frac{5 !}{3 !} \\
&=4 \times 5 \times 6-4 \times 5=100
\end{aligned}
\)
Together: The number of permutations of \(n\) different things taken all at a time when \(m\) specified things always come together is given as
\((n-m+1) ! \times m !\)Not-Together: The number of permutations of \(n\) different things taken all at a time when \(m\) specified things never come together is given as
\(n !-(n-m+1) ! \times m !\)
Example 5: In how many ways 6 children can be arranged in a line, such that
(a) Two particular children of them are always together
(b) Two particular children of them are never together
Solution:
(a) \((n-m+1) ! \times m ! =(6-2+1)! \times 2! = 5! \times 2! = 120 \times 2 = 240\)
Explanation: The given condition states that 2 students need to be together, hence we can consider them 1. Thus, the remaining 5 gives the arrangement in 5! ways, i.e. 120.
Also, the two children in a line can be arranged in 2! Ways.
Hence, the total number of arrangements will be,
\(5! \times 2 !=120 \times 2=240\) ways
(b) \(n !-(n-m+1) ! \times m ! = 6 ! – (6-2+1) ! \times 2 ! = 720 – 120 \times 2 = 480\)
Explanation: The total number of arrangements of 6 children will be 6 ! i.e. 720 ways.
Out of the total arrangement, we know that two particular children when together can be arranged in 240 ways. Therefore, the total arrangement of children in which two particular children are never together will be \(720-240\) ways, i.e. 480 ways.
Example 6: 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) \((n-m+1) ! \times m ! = (8-3+1) ! \times 3 ! = 4320\)
Explanation: 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 \({ }^6 \mathrm{P}_6=6\) !. 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 \(=6 ! \times 3 !=4320\).
(ii) \(n !-(n-m+1) ! \times m ! = 8 ! – (8-3+1) ! \times 3 ! = 8 ! – 6 ! \times 3 ! = 50 \times 720 = 36000\)
Explanation: 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
\(
\begin{aligned}
8 !-6 ! \times 3 ! &=6 !(7 \times 8-6) \\
&=2 \times 6 !(28-3) \\
&=50 \times 6 !=50 \times 720=36000
\end{aligned}
\)
Example 7: 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). Using the formula \(\frac{n !}{p_{1} ! p_{2} ! \ldots p_{k} !}\) we can find the number of arrangements = \(\frac{9 !}{4 ! 3 ! 2 !}=1260\).
Example 8: 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 \(\mathrm{N}\) appears 3 times, \(\mathbb{E}\) appears 4 times and \(\mathrm{D}\) appears 2 times and the rest are all different. Therefore The required number of arrangements \(=\frac{12 !}{3 ! 4 ! 2 !}=1663200\)
(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 \(P\)
\(
=\frac{11 !}{3 ! 2 ! 4 !}=138600 \text {. }
\)
(ii) There are 5 vowels in the given word, which are 4 Es and \(1 \mathrm{I}\). 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 \(3 \mathrm{Ns}\) and 2 Ds, can be rearranged in \(\frac{8 !}{3 ! 2 !}\) ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in \(\frac{5 !}{4 !}\) ways. Therefore, by multiplication principle, the required number of arrangements
\(
=\frac{8 !}{3 ! 2 !} \times \frac{5 !}{4 !}=16800
\)
(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 \(\mathrm{I}\) and \(\mathrm{P}\) at the extreme ends (I at the left end and \(\mathrm{P}\) at the right end). We are left with 10 letters.
Hence, the required number of arrangements
\(
=\frac{10 !}{3 ! 2 ! 4 !}=12600
\)
The number of permutations of \(n\) different things taken all at a time when \(m\) specified things are to be arranged such that no two of which are to occur together is \({ }^{n+1} P_m \times n\) !
The permutations of objects in a row are called linear permutations of linear arrangements. If \(n\) different things can be arranged in a row, the linear arrangement is \(n\) ! whereas every linear arrangement has a beginning and end, but in circular permutations, there is neither beginning nor end. When clockwise and anti-clockwise orders are taken as different, the number of circular permutations of \(n\) different things taken all at a time is \((n-1)\) !
But, when the clockwise and anti-clockwise orders are not different, i.e. the arrangements of beads in a necklace, arrangements of flowers in a garland, etc., the number of circular permutations of \(n\) different things \(=\frac{1}{2} \times(n-1) !\)
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