number of permutations by using kinds of duplicate elements

Each of the different arrangement which ccan be made by taking some or all of a number of things is called a permutation.
The number of ways of arranging n distinct object in a row taking r (0≤rn) at a time is denoted by P(n;r) or nPr.

Important results on Permutation

1. The number of permutations of n different things taken r at a time, allowing repetitions is nr.
2. The number of permutations of n different things taken r at a time is nPn=n!.
3. The number of permutations of n things taken all at a time, in which p are alike of one kind, q are alike of second kind an r are alike of thir kind and rest are different is

Permutations when any elements are not distinct

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, O1and 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 O1and O2are not treated as different, i.e., if O1and O2are the same O at both places.
Therefore, the required number of permutations =4!/2!=3×4=12.

Permutations when O1,O2 are different Permutations when O1,O2 are the same O.
RO1O2T
RO2O1T
–> ROOT
TO1O2R
TO2O1R
–> TOOR
RO1TO2
RO2TO1
–> ROTO
TO1RO2
TO2RO1
–> TORO
RTO1O2
RTO2O1
–> RTOO
TRO1O2
TRO2O1
–> TROO
O1O2RT
O2O1RT
–> OORT
O1RO2T
O2RO1T
–> OROT
O1TO2R
O2TO1R
–> OTOR
O1RTO2
O2RTO1
–> ORTO
O1TRO2
O2TRO1
–> OTRO
O1O2TR
O2O1TR
–> OOTR

Let us 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.
Temporarily, 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, I1NT1SI2T2UET3. Here if I1,I2are not same and T1,T2,T3are not same, then I1I2can be arranged in 2! ways and 4,T2,T3can 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!.
We can state (without proof) the following theorems:
Theorem 1: The number of permutations of n objects, where p objects are of the same kind and rest are all different =n!/em>p!.
In fact, we have a more general theorem.
Theorem 2: The number of permutations of n objects, where p1objects are of one kind, p2are of second kind, …, pkare of em>kth kind and the rest, if any, are of different kind is

Example 1: 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, 2L’s and rest are all different.
Therefore, the required number of arrangements

Example 2: 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 3: Find the number of arrangements of the letters of the word
INDEPENDENCE. In how many of these arrangements,
(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?
solutions: 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!2!4!)=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!/(3!2!) × 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 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
Example 4:
If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in list before the first word starting with E?
solution:
In the given word EXAMINATION, there are 11 letters out of which, A, land N appear 2 times and all the other letters appear only once.
The words that will be listed before the words starting with E in a dictionary will be the words that start with A only.
Therefore, to get the number of words starting with A, the letter A is fixed at the extreme left position, and then the remaining 10 letters taken all at a time are rearranged.
Since there are 2 Is and 2 Ns in the remaining 10 letters,
Number of words starting with A

=10!/2!2!=907200

Thus, the required numbers of words is 907200.

Example 5:
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
solution:
In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times, and M appears just once.
Therefore, number of distinct permutations of the letters in the given word

There are 4 Is in the given word. When they occur together, they are treated as single object [IIII] for the time being. This single object together with the remaining 7 objects will account for 8 objects.
These 8 objects in which there are 4 Ss and 2 Ps can be arranged in 8!/4!2! ways i.e., 840 ways.
Number of arrangements where all Is occur together =840.
Thus, number of distinct permutations of the letters in MISSISSIPPI in which four Is do not come together =34650-840=33810.
shorter solution:
Here, number of letters n=11. Number of ‘I’, p=4. Number of ‘S’, q=4. Number of ‘P’, r=2. Number of ‘M’, s=1
This is the formula for counting the number of permutations

If all 4 ‘I’ come together, they must be counted to one. So, number of permutations having all ‘I’ come together is equal to

Hence, required number of ways =34650-840=33810.

Example 6:
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together?
solution:
In the given word ASSASSINATION, the letter A appears 3 times, S appears 4 times, I appears 2 times, N appears 2 times, and all the other letters appear only once. Since all the words have to be arranged in such a way that all the Ss are together, SSSS is treated as a single object for
the time being. This single object together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 Is, and 2 Ns can be arranged in 10!/(3!2!2!)
ways. Thus, Required number of ways of arranging the letters of the given word =10!/(3!2!2!)=151200.

Example 7:
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(ii) Vowels are all together,
(iii) There are always 4 letters between P and S?
solutions:
In the word PERMUTATIONS, there are 2 Ts and all the other letters appear only once.
(i) If P and S are fixed at the extreme ends (P at the left end and S at the right end), then 10 letters are left.
Hence, in this case, required number of arrangements

=10!/2!=1814400

(ii) There are 5 vowels in the given word, each appearing only once.
Since they have to always occur together, they are treated as a single object for the time being.
This single object together with the remaining 7 objects will account for 8 objects.
These 8 objects in which there are 2 Ts can be arranged in 8!/2! ways.
Corresponding to each of these arrangements, the 5 different vowels can be arranged in 5! ways.
Therefore, by multiplication principle, required number of arrangements in this case

=8!/(2!×5!)=2419200

(iii) The letters have to be arranged in such a way that there are always 4 letters between P and S.
Therefore, in a way, the places of P and S are fixed. The remaining 10 letters in which there are 2 Ts can be arranged in 10!/2! ways.
Also, the letters P and S can be placed such that there are 4 letters between them in 2×7=14 ways.
Therefore, by multiplication principle, required number of arrangements in this case

=10!/(2!×14!)=25401600.