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 \({n}\) distinct objects, taken all at a time, such that \({m}\) specific objects are arranged so that no two of them occur together is given by \({ }^{n-m+1} P_m \times(n-m)!\)
Correct Reasoning (Gap Method):
Step 1: Arrange the \(n-m\) non-special objects
They are all distinct, so they can be arranged in:
\(
(n-m)!
\)
This arrangement creates gaps where the \(m\) special objects may be placed. For \(k\) objects, the number of gaps is:
\(
k+1
\)
So the \((n-m)!\) arrangement creates:
\(
(n-m)+1=n-m+1 \quad \text { gaps }
\)
Step 2: Place the \(m\) special objects
Each of the \(m\) special objects must go into distinct gaps (to avoid adjacency).
Number of ways to choose and order the \(m\) gaps:
\(
{ }^{n-m+1} P_m
\)
(We are permuting \(m\) distinct items into \(n-m+1\) available gaps.)
So the correct formula is: \({ }^{n-m+1} P_m \times(n-m)!\)
For example, suppose we have:
\(n=6\) total distinct objects, \(\mathrm{m}=2\) specific objects that must not occur together
Let the special objects be A and B , and the others be 1, 2, 3, 4.
Step 1: Arrange the non-special objects
The non-special objects are: \(1,2,3,4 \rightarrow 4\) objects
A typical arrangement looks like:
\(
1342
\)
This creates gaps:
\(
\_1 \_3 \_4 \_2 \_
\)
There are:
\(
(n-m+1)=4+1=5 \text { gaps }
\)
Step 2: Place A and B in the gaps (not together)
We have \(\mathbf{5}\) gaps and must place \(\mathbf{2}\) distinct special objects in separate gaps.
Number of ways:
\(
{ }^5 P_2=5 \times 4=20
\)
Because:
choose a gap for A (5 choices)
choose another gap for B (4 choices)
A and B are distinguishable → order matters
Step 3: Multiply both steps
\(
\begin{aligned}
& \text { Total }={ }^5 P_2 \times 4! \\
& =20 \times 24=480
\end{aligned}
\)
The number of permutations of 6 distinct objects where \(A\) and \(B\) never occur together is: 480.
Example 9: Find the number of ways to arrange the letters in the word “SIGNATURE” such that no two vowels are together.
Solution: Total items ( \(n\) ): 9 letters (S, I, G, N, A, T, U, R, E)
Specified items to be separated \((m): 4\) vowels (I, A, U, E)
Unrestricted items: 5 consonants (S, G, N, T, R)
Step 1: Arrange the consonants
The 5 consonants can be arranged in 5 ! ways.
\(
5!=5 \times 4 \times 3 \times 2 \times 1=120
\)
Step 2: Place the vowels in the gaps
Arranging 5 consonants creates \(5+1=6\) gaps:
\(
_{-} \mathrm{S}_{-} \mathrm{G}_{-} \mathrm{N}_{-} \mathrm{T}_{-} \mathrm{R}_{-}
\)
We need to place the 4 vowels into 4 of these 6 gaps such that no two are together. The number of ways to select and arrange them in the gaps is \(P(6,4)\) or \({ }^6 P_4\).
\(
P(6,4)=\frac{6!}{(6-4)!}=\frac{6!}{2!}=\frac{720}{2}=360
\)
The total number of arrangements is the product of the ways for each step:
\(
\text { Total ways }=120 \times 360=43,200
\)
Using the general formula:
\(
P(n-m+1, m) \times(n-m)!=P(9-4+1,4) \times(9-4)!=P(6,4) \times 5!=43,200.
\)
Circular Permutations
In a linear permutation, an arrangement has a definite beginning and end, so for \(n\) distinct objects:
Number of linear arrangements \(=n!\)
But in a circular arrangement, there is no fixed beginning or end.
To avoid overcounting, we fix one object as a reference point.
Case 1: Clockwise and anticlockwise are considered different
This is the usual case for seating people around a table.
Fix one object (e.g., seat one person at a point).
Arrange the remaining \(n-1\) distinct objects around the circle.
\(
(n-1)!
\)
Case 2: Clockwise and anticlockwise arrangements are considered the SAME
This applies to symmetric objects like:
(i) beads in a necklace
(ii) flowers in a garland
(iii) colored flags on a circular string
A clockwise arrangement and its mirror (anticlockwise) are identical.
So each circular arrangement is counted twice in case 1.
Therefore:
\(
\text { Number of distinct circular arrangements }=\frac{1}{2} \times(n-1)!
\)
Final Summary:
\(
\begin{array}{ll}
\hline \text { Type of Circular Arrangement } & \text { Formula } \\
\hline \text { Clockwise and anticlockwise different } & (n-1)! \\
\hline \text { Clockwise and anticlockwise not different } & \frac{1}{2}(n-1)! \\
\hline
\end{array}
\)
Example 10: Clockwise and Anticlockwise ARE Different:
3 people (A, B, C) seated around a round table
Solution: Formula:
\(
(n-1)!=(3-1)!=2!=2
\)
Enumerating the arrangements
Fix \(\mathbf{A}\) at the top position (to remove rotation symmetry).
Now arrange B and C around the circle:
1. \(\mathrm{A}-\mathrm{B}-\mathrm{C}\)
2. \(\mathrm{A}-\mathrm{C}-\mathrm{B}\)
Both are counted as different because clockwise \(\neq\) anticlockwise.
Example 11: Clockwise and Anticlockwise ARE the Same (Necklace / Garland):
4 beads (A, B, C, D) arranged on a necklace.
Solution: Formula:
\(
\frac{1}{2}(n-1)!=\frac{1}{2}(4-1)!=\frac{1}{2} \times 3!=\frac{6}{2}=3
\)
Why only 3 arrangements?
Fix \(A\) at the top.
Arrange B, C, D around the circle:
ABCD
ABDC
ACBD
ACDB
ADBC
ADCB
(6 arrangements = the usual ( \(n-1\) )!)
But each arrangement has a mirror image that looks the same on a necklace, so we divide by 2:
Pairs of symmetric arrangements:
\(\mathrm{ABCD} \leftrightarrow \mathrm{ADCB}\)
\(\mathrm{ABDC} \leftrightarrow \mathrm{ACDB}\)
\(\mathrm{ACBD} \leftrightarrow \mathrm{ADBC}\)
Thus, only 3 unique necklaces.
Example 12: 5 children sitting in a circle
Solution: Since clockwise \(\neq\) anticlockwise for sitting:
\(
(5-1)!=4!=24
\)
There are 24 distinct ways to seat 5 children around a table.
Example 13: 6 flowers on a garland
Solution: Clockwise = anticlockwise (because a garland can be flipped):
\(
\frac{1}{2}(6-1)!=\frac{1}{2} \times 5!=\frac{1}{2} \times 120=60
\)
There are 60 distinct garland arrangements.
Example 14: 7 distinct flags around a circular rope (rotations matter, flips matter too)
Solution: Clockwise \(\neq\) anticlockwise (rope is fixed, cannot be flipped):
\(
(7-1)!=6!=720
\)
There are \(\mathbf{7 2 0}\) distinct ways.
Example 15: Circular permutation with repeated objects
Arrange 6 beads: A, A, B, C, D, E on a necklace
Solution: Given:
It is a necklace → clockwise and anticlockwise are the same.
There are repeated objects (two A’s).
Step 1: Circular permutations with rotation only
If all beads were distinct:
\(
(6-1)!=5!=120
\)
But two A’s are identical, so divide by 2!:
\(
\frac{5!}{2!}=\frac{120}{2}=60
\)
Step 2: Account for reflection (mirror symmetry)
Because it is a necklace:
\(
\text { Final count }=\frac{1}{2} \times \frac{5!}{2!}=\frac{1}{2} \times 60=30
\)
Answer: 30 distinct necklace arrangements.
Example 16: 8 people sit around a table; A and B must sit together
Solution: This is a seating (table) problem, so clockwise \(\neq\) anticlockwise:
Step 1: Treat A-B as one block
So instead of 8 people, we have:
1 block (AB)
6 others
Total = 7 “objects” in a circle:
\(
(7-1)!=6!=720
\)
Step 2: Arrange A and B inside the block
\(
2!=2
\)
Final answer:
\(
720 \times 2=1440
\)
Example 17: 10 people around a table; 3 specific people must NOT sit adjacent.
Solution: Total arrangements: \((10-1)!=9!=362880\)
We subtract arrangements where at least two of the 3 sit together.
This is a classic Inclusion-Exclusion problem.
Let the three be A, B, C.
Case 1: Exactly 2 sit together
Example: A and B together, C separate.
Treat AB as a block → now 9 objects ( \(\mathrm{AB}+8\) others):
\((9-1)!=8!=40320\)
Internal arrangements:
\(A B, B A \rightarrow 2!\)
So: \(40320 \times 2=80640\)
This happens for pairs (A,B), (A,C), (B,C):
\(
3 \times 80640=241920
\)
Case 2: All 3 sit together
ABC block (3! internal).
Total objects \(=8\) :
\((8-1)!=7!=5040\)
Internal arrangements:
\(3!=6\)
Total:
\(5040 \times 6=30240\)
Subtract this (because it was counted in the pair cases):
\(\text { Forbidden }=241920-30240=211680\)
Final allowed arrangements
\(9!-211680=362880-211680=151200\)
Case I: Clockwise and anticlockwise arrangements are considered DIFFERENT
If clockwise and anti-clockwise arrangements are taken as different, the number of circular permutations of \(n\) other things, taken \(r\) at a time is given by \(\frac{{ }^n P_r}{r}\)
Reasoning: When you take \({r}\) objects out of \({n}\) and arrange them around a circle, some positions overlap because rotations (and possibly reflections) are considered identical. This is not the same as taking all \(n\)-instead, you choose r out of n and then arrange them in a circle.
Step 1: Choose and arrange \({r}\) objects in a row
Number of linear permutations of \({r}\) objects chosen from \({n}\) :
\(
{ }^n P_r=\frac{n!}{(n-r)!}
\)
Step 2: Remove rotational duplicates
When arranging \(r\) objects in a circle, rotating the circle by \(1,2, \ldots, \mathrm{r}\) positions gives the same circular arrangement.
So each distinct circle corresponds to \({r}\) linear arrangements.
Thus:
\(
\text { Circular permutations }=\frac{{ }^n P_r}{r}
\)
Final Formula (clockwise \(\neq\) anticlockwise): \(\frac{{ }^n P_r}{r}\)
Case 2: Clockwise and anticlockwise arrangements are considered the SAME
If clockwise and anti-clockwise arrangements are not taken as different, the number of circular permutations of \(n\) other things, taken \(r\) at a time is \(\frac{{ }^n P_r}{2 r}\)
Reasoning: (Necklace, garland, bracelet, beads, etc.)
In addition to the \({r}\) rotations, each arrangement also has a mirror image that is considered identical.
Therefore, you divide by \(2 r\) instead of \(r\).
Circular permutations \(=\frac{{ }^n P_r}{2 r}\)
Final Formula (clockwise = anticlockwise): \(\frac{{ }^n P_r}{2 r}\)
Example 18: (Advanced): From 12 beads choose 5 to make a bracelet.
Solution: Bracelet → flip symmetry → clockwise \(=\) anticlockwise
\(
\begin{gathered}
n=12, r=5 \\
\frac{12 P_5}{2 \cdot 5}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{10}=\frac{95040}{10}=9504
\end{gathered}
\)
Example 19: (Tricky): 4 musicians selected from 8 sit in a circle; two opposite directions count the same.
Solution:
→ They sit in a circle but orientation doesn’t matter
→ clockwise = anticlockwise
\(
\begin{gathered}
n=8, r=4 \\
\frac{{ }^8 P_4}{8}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{8}=7 \cdot 6 \cdot 5=210
\end{gathered}
\)
(Notice: dividing by \(2 r=8\).)
Example 20: (Very Advanced): 7 people selected from 10, arranged in a circle; two arrangements that differ by rotation or flip are identical.
Solution:
→ both rotation and reflection symmetries
→ clockwise \(=\) anticlockwise
\(
\begin{gathered}
n=10, r=7 \\
\frac{10 P_7}{2 \cdot 7}=\frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{14} \\
=\frac{604800}{14}=43200
\end{gathered}
\)
Example 21: The number of nine-non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is
a. 2(4!)
b. \(3(7!) / 2\)
c. \(2(7!)\)
d. \({ }^4 P_4 \times{ }^4 P_4\)
Solution: (d) Step 1: Understand the constraints and identify the middle digit
The problem requires forming a nine-digit number using all non-zero digits (1 through 9) exactly once. The key constraint is the middle digit ( \(d_5\) ). The first four digits must be less than \(d_5\), and the last four digits must be greater than \(d_5\). This means the set of digits \(\{1,2, \ldots, 9\}\) is split into three parts: a set of 4 smaller digits, the middle digit, and a set of 4 larger digits.
This condition is only met if the number of digits less than \(d_5\) is 4 and the number of digits greater than \(d_5\) is 4. The only digit in the range \([1,9]\) that satisfies this is \(d_5=5\). The digits less than 5 are \(\{1,2,3,4\}\), and the digits greater than 5 are \(\{6,7,8,9\}\).
Step 2: Calculate the number of arrangements for the first four places
The first four places must be filled using the digits \(\{1,2,3,4\}\) without repetition, in any order. The number of permutations of 4 distinct items in 4 places is given by \(4!\) or \({ }^{\mathbf{4}} P_{\mathbf{4}}\).
\(
{ }^4 P_4=4!=4 \times 3 \times 2 \times 1=24
\)
Step 3: Calculate the number of arrangements for the last four places
The last four places must be filled using the digits \(\{6,7,8,9\}\) without repetition, in any order. The number of permutations of 4 distinct items in 4 places is also given by \(4!\) or \({ }^4 P_4\).
\(
{ }^4 P_4=4!=4 \times 3 \times 2 \times 1=24
\)
Step 4: Calculate the total number of such nine-digit numbers
The total number of possible nine-digit numbers is the product of the number of choices for each section: one choice for the middle digit, 24 arrangements for the first four digits, and 24 arrangements for the last four digits.
\(
\text { Total Ways }=1 \times{ }^4 P_4 \times{ }^4 P_4=4!\times 4!=24 \times 24=576
\)
The number of nine-non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is \({ }^{\mathbf{4}} \boldsymbol{P}_{\mathbf{4}} \times{ }^{\mathbf{4}} \boldsymbol{P}_{\mathbf{4}}\).
Example 22: If all the permutations of the letters in the word ‘OBJECT’ are arranged (and numbered serially) in alphabetical order as in a dictionary, then the \(717^{\text {th }}\) word is
a. TOJECB
b. TOEJBC
c. TOCJEB
d. TOJCBE
Solution: (d) The order of letters of the word ‘OBJECT’ is B C E J O T Words starting with B can be formed in 5 ! ways. Words starting with C can be formed in 5 ! ways. Words starting with E can be formed in 5 ! ways. Words starting with J can be formed in 5 ! ways. Words starting with O can be formed in 5 ! ways. Words starting with TB can be formed in 4 ! ways. Words starting with TC can be formed in 4! ways. Words starting with TE can be formed in 4! ways. Words starting with TJ can be formed in 4! ways. Words starting with TOB can be formed in 3! ways. Words starting with TOC can be formed in 3! ways. Words starting with TOE can be formed in 3! ways. Words starting with TOJB can be formed in 2! ways. Words starting with TOJC can be formed in 2! ways. Therefore, the total number of words is 718 words. Hence \(717^{\text {th }}\) word is TOJCBE.
Example 23: In a three-storey building, there are four rooms on the ground floor, two on the first and two on the second floor. If the rooms are to be allotted to six persons, one person occupying one room only, the number of ways in which this can be done so that no floor remains empty is
a. \({ }^8 P_6-2(6!)\)
b. \({ }^8 P_6\)
c. \({ }^8 P_5(6!)\)
d. none of these
Solution: (a) The number of ways of allotment without any restriction is \({ }^8 P_6\). Now, it is possible that all rooms of \(2^{\text {nd }}\) floor or \(3^{\text {rd }}\) floor are not occupied. Thus, there are two ways in which one floor remains unoccupied. Hence, the number of ways of allotment in which a floor is unoccupied is \(2 \times 6\) ! Hence, number of ways in which none of the floor remains unoccupied is \({ }^8 P_6- 2(6!)\).
Example 24: The total number of three-letter words that can be formed from the letter of the word ‘SAHARANPUR’ is equal to
a. 210
b. 237
c. 247
d. 227
Solution: (c) Let’s compute the number of 3-letter words from SAHARANPUR.
Letters and frequencies:
\(\mathrm{A}=3\)
\(\mathrm{R}=2\)
\(\mathrm{S}, \mathrm{H}, \mathrm{N}, \mathrm{P}, \mathrm{U}=1\) each
Total distinct letters \(=7\).
Case 1: All letters different
Choose 3 distinct letters from \(7 \rightarrow\binom{7}{3}=35\)
Each can be arranged in \(3!=6\) ways
Total \(=35 \times 6=210\)
Case 2: Exactly two letters the same
Only \(A\) and \(R\) can repeat.
AAx: choose the third letter from 6 choices \(\rightarrow 6\) words
Each arrangement: \(3!/ 2!=3\)
Total \(=6 \times 3=18\)
RRx: choose third letter from 6 choices \(\rightarrow 6\) words
Arrangements = 3
Total \(=6 \times 3=18\)
Total for this case \(=18+18=36\)
Case 3: All three letters identical
Only AAA possible → 1 word
Grand Total
\(
210+36+1=247
\)
Example 25: The number of words of four letters that can be formed from the letters of the word ‘EXAMINATION’ is
a. 1464
b. 2454
c. 1678
d. none of these
Solution: (b) All four letters are different: There are 8 distinct letters (E, X, A, M, I, N, T, O). The number of arrangements is \({ }^8 P_4\) or \({ }^8 C_4 \times 4\) !, which equals 1680.
One letter is duplicated, the other two distinct: There are 3 letters that appear twice (A, I, N).
Choose one letter to duplicate ( \({ }^3 C_1\) ways) and two distinct letters from the remaining seven ( \({ }^7 C_2\) ways).
Arrange these 4 letters ( 2 same, 2 distinct) in \(4!/ 2\) ! ways.
Total for this case: \({ }^3 C_1 \times{ }^7 C_2 \times(4!/ 2!)=3 \times 21 \times 12=756\).
Two different letters are duplicated (two pairs of same letters):
Choose two letters to duplicate from the three available ( \({ }^3 C_2\) ways).
Arrange these 4 letters ( 2 same, 2 same) in \(4!/(2!2!)\) ways.
Total for this case: \({ }^3 C_2 \times(4!/(2!2!))=3 \times 6=18\).
The total number of words is the sum of these cases: \(1680+756+18=2454\).
You cannot copy content of this page