CBSE Class 11 Applied Mathematics Permutations and Combinations Notes Set 01

Factorial

The Factorial

The continued product of the first n natural numbers is called 'n factorial' and is denoted by n!
i.e. n! = 1 × 2 × 3 × … × (n − 1) × n
Thus 3! = 1 × 2 × 3 = 6
Clearly, n! is defined for positive integers only.
Zero factorial: It does not make any sense to define it as the product of the integers from 1 to zero. So, we define 0! = 1.
Note: Factorials of proper fractions or negative integers are not defined.
Deduction: n! = 1 × 2 × 3 × … × (n − 1) × n = [1 × 2 × 3 × … × (n − 1)] × n = (n − 1)! × n
∴ n! = n(n − 1)!

Example 1. Convert the following products into factorials: (i) 6 · 7 · 8 · 9 · 10 (ii) 2 · 4 · 6 · 8 · 10
Answer:
(i) \( 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 = \frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{1 \times 2 \times 3 \times 4 \times 5} = \frac{10!}{5!} \)
(ii) 2 · 4 · 6 · 8 · 10 = (2 × 1) × (2 × 2) × (2 × 3) × (2 × 4) × (2 × 5) = 2⁵ × 1 × 2 × 3 × 4 × 5 = 2⁵ × 5!

Example 2. Find the LCM of 4!, 5! and 6!
Answer:
We have 5! = 5 × 4! and 6! = 6 × 5 × 4!
∴ LCM of 4!, 5!, 6! = LCM [4!, 5 × 4!, 6 × 5 × 4!] = 4! × 5 × 6 = 6! = 720

Example 3. Prove that (n! + 1) is not divisible by any natural number between 2 and n.
Answer:
If m is divisible by k and r is any natural number less than k, then on dividing m + r by k we obtain r as the remainder.
Since n! = 1 · 2 · 3 · … · (n − 1) · n, it follows that n! is divisible by every natural number between 2 and n.
So (n! + 1), when divided by any natural number between 2 and n, leaves 1 as the remainder.
Hence, (n! + 1) is not divisible by any natural number between 2 and n.

Fundamental Principles of Counting

Fundamental Principle of Multiplication

If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, the second job can be completed in n ways, then the two jobs in succession can be completed in m × n ways.

Example 4. In a class there are 10 boys and 8 girls. The teacher wants to select a boy and a girl to represent the class in a function. In how many ways can the teacher make this selection?
Answer:
Here the teacher is to perform two jobs: (i) selecting a boy among 10 boys, and (ii) selecting a girl among 8 girls.
The first of these can be performed in 10 ways and the second in 8 ways.
Therefore, by the fundamental principle of multiplication, the required number of ways is 10 × 8 = 80.

Fundamental Principle of Addition

If there are two jobs such that they can be performed independently in m and n ways respectively, then either of the two jobs can be performed in (m + n) ways.

Example 5. In a class there are 10 boys and 8 girls. The teacher wants to select either a boy or a girl to represent the class in a function. In how many ways can the teacher make this selection?
Answer:
Here the teacher is to perform either of the following two jobs: (i) selecting a boy among 10 boys, or (ii) selecting a girl among 8 girls.
The first of these can be performed in 10 ways and the second in 8 ways. Therefore, by the fundamental principle of addition, either of the two jobs can be performed in (10 + 8) = 18 ways.
Hence the teacher can make the selection of either a boy or a girl in 18 ways.

Example 6. In a monthly test, the teacher decides that there will be three questions, one from each of Exercises 7, 8 and 9 of the textbook. If there are 12 questions in Exercise 7, 18 in Exercise 8 and 9 in Exercise 9, in how many ways can three questions be selected?
Answer:
There are 12 questions in Exercise 7, so one question from Exercise 7 can be selected in 12 ways.
Exercise 8 contains 18 questions, so the second question can be selected in 18 ways. There are 9 questions in Exercise 9, so the third question can be selected in 9 ways.
Hence, three questions can be selected in 12 × 18 × 9 = 1944 ways.

Example 7. How many three-digit numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6?
Answer:
We have to determine the total number of three-digit numbers formed by using the digits 1, 7, 8, 9. Clearly, the repetition of digits is allowed.
A three-digit number has three places, i.e. units, tens and hundreds. The units place can be filled by any of the digits 1, 7, 8, 9, so it can be filled in 4 ways. Similarly, each of the tens and hundreds places can be filled in 4 ways.
∴ Total number of required numbers = 4 × 4 × 4 = 64

Example 8. How many numbers are there between 100 and 1000 in which all the digits are distinct?
Answer:
A number between 100 and 1000 has three digits. So we have to form all possible 3-digit numbers with distinct digits.
We cannot have 0 at the hundreds place. So the hundreds place can be filled with any of the 9 digits 1, 2, …, 9, i.e. in 9 ways.
Now 9 digits are left, including 0. So the tens place can be filled with any of the remaining 9 digits in 9 ways. Then the units place can be filled with any of the remaining 8 digits in 8 ways.
Hence, the total number of required numbers = 9 × 9 × 8 = 648.

Example 9. How many numbers are there between 100 and 1000 such that 7 is in the units place?
Answer:
Every number between 100 and 1000 is a three-digit number. So we have to form 3-digit numbers with 7 at the units place by using the digits 0, 1, 2, …, 9. Clearly, repetition of digits is allowed.
The hundreds place can be filled with any of the digits from 1 to 9, i.e. in 9 ways. The tens place can be filled with any of the digits from 0 to 9, i.e. in 10 ways. Since all the numbers have 7 at the units place, the units place can be filled in only one way.
Hence, by the fundamental principle of counting, the total number of numbers between 100 and 1000 having 7 at the units place = 9 × 10 × 1 = 90.

Example 10. How many numbers are there between 100 and 1000 such that at least one of their digits is 7?
Answer:
Clearly, a number between 100 and 1000 has 3 digits.
Number of 3-digit numbers having at least one digit as 7 = (Total number of 3-digit numbers) − (Number of 3-digit numbers in which 7 does not appear at all)
Total number of 3-digit numbers: We form 3-digit numbers by using the digits 0, 1, 2, …, 9. The hundreds place can be filled in 9 ways and each of the tens and units places in 10 ways. So the total number of 3-digit numbers = 9 × 10 × 10 = 900.
3-digit numbers in which 7 does not appear at all: We use the digits 0 to 9 except 7. The hundreds place can be filled in 8 ways and each of the tens and units places in 9 ways. So their number = 8 × 9 × 9 = 648.
Hence, the total number of 3-digit numbers having at least one of their digits as 7 = 900 − 648 = 252.

Example 11. How many numbers are there between 100 and 1000 which have exactly one of their digits as 7?
Answer:
A number between 100 and 1000 contains 3 digits. Numbers having exactly one of their digits as 7 can be divided into three types:
(i) Those that have 7 in the units place but not in any other place: there are 8 × 9 × 1 = 72 such numbers.
(ii) Those that have 7 in the tens place but not in any other place: there are 8 × 1 × 9 = 72 such numbers.
(iii) Those that have 7 in the hundreds place but not in any other place: there are 1 × 9 × 9 = 81 such numbers.
Hence, the total number of required numbers = 72 + 72 + 81 = 225.

Example 12. A gentleman has 6 friends to invite. In how many ways can he send invitation cards to them, if he has three servants to carry the cards?
Answer:
Since a card can be sent by any one of the three servants, the number of ways of sending the invitation card to the first friend = 3.
Similarly, the invitation card can be sent to each of the six friends in 3 ways.
So the required number of ways = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶ = 729.

Example 13. How many three-digit odd numbers can be formed by using the digits 1, 2, 3, 4, 5, 6 if (i) the repetition of digits is not allowed? (ii) the repetition of digits is allowed?
Answer:
For a number to be odd, we must have 1, 3 or 5 at the units place. So there are 3 ways of filling the units place.
(i) Since repetition of digits is not allowed, the tens place can be filled with any of the remaining 5 digits in 5 ways. Now four digits are left, so the hundreds place can be filled in 4 ways. So the required number of numbers = 3 × 5 × 4 = 60.
(ii) Since repetition of digits is allowed, each of the tens and hundreds places can be filled in 6 ways. Hence the required number of numbers = 3 × 6 × 6 = 108.

Example 14. In how many ways can 5 rings of different types be worn in 4 fingers?
Answer:
The first ring can be worn in any of the 4 fingers. So there are 4 ways of wearing it.
Similarly, each one of the other rings can be worn in 4 ways.
Hence, the required number of ways = 4 × 4 × 4 × 4 × 4 = 4⁵ = 1024.

Example 15. By using the digits 0, 1, 2, 3, 4 and 5 (repetitions not allowed), numbers are formed by using any number of digits. Find the total number of non-zero numbers that can be formed.
Answer:
Number of numbers = number of 1-digit numbers + number of 2-digit numbers + … + number of 6-digit numbers
(For each length the first place cannot be 0, so it can be filled in 5 ways; the next places in 5, 4, 3, 2, 1 ways.)
= 5 + 5 × 5 + 5 × 5 × 4 + 5 × 5 × 4 × 3 + 5 × 5 × 4 × 3 × 2 + 5 × 5 × 4 × 3 × 2 × 1
= 5 + 25 + 100 + 300 + 600 + 600 = 1630

Permutations

Permutation

Each of the arrangements which can be made by taking some or all of a number of things is called a permutation.
Example: If there are three objects a, b, c, then the permutations of these objects taken two at a time are ab, ba, bc, cb, ac, ca. So the number of permutations of three different things taken two at a time is 6.
Notation: If n and r are positive integers such that 1 ≤ r ≤ n, then the number of all permutations of n distinct things taken r at a time is denoted by the symbol P(n, r) or ⁿP_r.
Thus ⁿP_r or P(n, r) = total number of permutations of n distinct things, taken r at a time.

Formulae for ⁿP_r

Theorem 1: Let r and n be positive integers such that 1 ≤ r ≤ n. Then the number of all permutations of n distinct things taken r at a time is given by P(n, r) = ⁿP_r = n(n − 1)(n − 2) … {n − (r − 1)}.
Theorem 2: \( P(n, r) = {}^{n}P_{r} = \frac{n!}{(n - r)!} \)
Proof: \( {}^{n}P_{r} = n(n - 1)(n - 2) \ldots (n - (r - 1)) = \frac{n(n - 1)(n - 2) \ldots (n - (r - 1))(n - r)(n - (r + 1)) \ldots 3 \cdot 2 \cdot 1}{(n - r)(n - (r + 1)) \ldots 3 \cdot 2 \cdot 1} = \frac{n!}{(n - r)!} \)
Theorem 3: The number of all permutations of n distinct things, taken all at a time, is n!
Proof: P(n, n) = n(n − 1)(n − 2) … (n − (n − 1)) = n(n − 1)(n − 2) … 3 · 2 · 1 = n!
Theorem 4: 0! = 1
Proof: We have \( {}^{n}P_{r} = \frac{n!}{(n - r)!} \). Putting r = n, \( {}^{n}P_{n} = \frac{n!}{(n - n)!} \) ⇒ \( n! = \frac{n!}{0!} \) ⇒ \( 0! = \frac{n!}{n!} = 1 \)

Example 16. In how many ways can three different rings be worn in four fingers with at most one in each finger?
Answer:
The total number of ways is the same as the number of arrangements of 4 fingers, taken 3 at a time. So the required number of ways \( = {}^{4}P_{3} = \frac{4!}{(4 - 3)!} = \frac{4!}{1!} = 24 \)
Aliter: Let R₁, R₂, R₃ be the three rings. R₁ can be put in any one of the four fingers, so there are 4 ways in which R₁ can be worn. Now R₂ can be worn in any one of the remaining three fingers in 3 ways. In the remaining 2 fingers, R₃ can be worn in 2 ways.
So, by the fundamental principle of counting, the total number of ways = 4 × 3 × 2 = 24.

Example 17. Seven athletes are participating in a race. In how many ways can the first three prizes be won?
Answer:
The total number of ways in which the first three prizes can be won is the number of arrangements of seven different things taken 3 at a time.
So the required number of ways \( = {}^{7}P_{3} = \frac{7!}{(7 - 3)!} = \frac{7!}{4!} = 7 \times 6 \times 5 = 210 \)
Aliter: The first prize can be won in 7 ways. The second prize can be won by any one of the remaining six athletes in 6 ways. Five athletes are left, so the third prize can be won in 5 ways. Hence the required number of ways = 7 × 6 × 5 = 210.

Example 18. In how many ways can 6 persons stand in a queue?
Answer:
The number of ways in which 6 persons can stand in a queue is the same as the number of arrangements of 6 different things taken all at a time.
Hence the required number of ways = ⁶P₆ = 6! = 720.

Example 19. It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Answer:
In all, 9 persons are to be seated in a row, and in the row of 9 positions there are exactly four even places, viz. the second, fourth, sixth and eighth.
These four even places are to be occupied by the 4 women. This can be done in ⁴P₄ ways. The remaining 5 positions can be filled by the 5 men in ⁵P₅ ways.
So, by the fundamental principle of counting, the number of seating arrangements = ⁴P₄ × ⁵P₅ = 4! × 5! = 24 × 120 = 2880.

Example 20. How many different signals can be given using any number of flags from 5 flags of different colours?
Answer:
The signals can be made by using one or two or three or four or five flags at a time.
Hence, by the fundamental principle of counting, the total number of signals = ⁵P₁ + ⁵P₂ + ⁵P₃ + ⁵P₄ + ⁵P₅
= 5 + 5 × 4 + 5 × 4 × 3 + 5 × 4 × 3 × 2 + 5 × 4 × 3 × 2 × 1 = 5 + 20 + 60 + 120 + 120 = 325

Example 21. How many numbers lying between 100 and 1000 can be formed with the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
Answer:
Every number lying between 100 and 1000 is a three-digit number. Therefore, we have to find the number of permutations of the five digits 1, 2, 3, 4, 5 taken three at a time.
Hence, the required number of numbers \( = {}^{5}P_{3} = \frac{5!}{(5 - 3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60 \)

Example 22. How many four-digit numbers are there with distinct digits?
Answer:
The total number of arrangements of the ten digits 0, 1, 2, …, 9 taking 4 at a time is ¹⁰P₄. But these arrangements also include those which have 0 at the thousands place. Such numbers are not four-digit numbers.
When 0 is fixed at the thousands place, we have to arrange the remaining 9 digits taking 3 at a time. The number of such arrangements is ⁹P₃.
Hence, the total number of four-digit numbers = ¹⁰P₄ − ⁹P₃ = 5040 − 504 = 4536.

Example 23. Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Determine the number of words which have at least one letter repeated.
Answer:
The number of 5-letter words which can be formed from 10 letters when repetition is allowed = 10 × 10 × 10 × 10 × 10 = 10⁵
The number of 5-letter words which can be formed when none of their letters is repeated = number of arrangements of 10 letters taking 5 at a time = ¹⁰P₅ = 30240
Hence, the number of 5-letter words which have at least one of their letters repeated = 10⁵ − 30240 = 69760.

Example 24. In how many ways can 7 pictures be hung from 5 picture nails on a wall?
Answer:
The number of ways in which 7 pictures can be hung from 5 picture nails on a wall is the same as the number of arrangements of 7 things, taking 5 at a time.
Hence the required number \( = {}^{7}P_{5} = \frac{7!}{(7 - 5)!} = \frac{7!}{2!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520 \)

Example 25. There are six periods in each working day of a school. In how many ways can one arrange 5 subjects such that each subject is allotted at least one period?
Answer:
Since there are 6 periods and 5 subjects, exactly one subject gets two periods and each of the other four subjects gets one period.
The subject which gets two periods can be chosen in 5 ways.
The six periods are then filled by 6 subjects of which two are alike, which can be done in \( \frac{6!}{2!} = 360 \) ways.
Hence, the total number of arrangements = 5 × 360 = 1800.

Permutations under Certain Conditions

Theorem 1: The number of all permutations of n different objects taken r at a time, when a particular object is always to be included in each arrangement, is r · ⁿ⁻¹P_r−1.
Theorem 2: The number of permutations of n distinct objects taken r at a time, when a particular object is never taken in each arrangement, is ⁿ⁻¹P_r.
Theorem 3: The number of permutations of n different objects taken r at a time, in which two specified objects always occur together, is 2!(r − 1) · ⁿ⁻²P_r−2.

Example 26. In how many ways can the letters of the word PENCIL be arranged so that (i) N is always next to E (ii) N and E are always together?
Answer:
(i) Let us keep EN together and consider it as one letter. Now we have 5 letters, which can be arranged in a row in ⁵P₅ = 5! = 120 ways. Hence the total number of ways in which N is always next to E is 120.
(ii) Keeping E and N together and considering them as one letter, we have 5 letters which can be arranged in ⁵P₅ = 5! ways. But E and N can be put together in 2! ways (viz. EN, NE).
Hence the total number of ways = 5! × 2! = 240.

Example 27. How many different words can be formed with the letters of the word EQUATION so that the words begin and end with a consonant?
Answer:
There are 3 consonants (Q, T, N), and all words should begin and end with a consonant. So the first and last places can be filled with the 3 consonants in ³P₂ ways.
Now the remaining 6 places are to be filled up with the remaining 6 letters in ⁶P₆ ways.
Hence the required number of words = ³P₂ × ⁶P₆ = 6 × 720 = 4320.

Example 28. In how many ways can 5 boys and 3 girls be seated in a row so that no two girls are together?
Answer:
The 5 boys can be seated in a row in ⁵P₅ = 5! ways. In each of these arrangements, 6 places are created, shown by the cross marks, as given below:
× B × B × B × B × B ×
Since no two girls are to sit together, we may arrange the 3 girls in these 6 places. This can be done in ⁶P₃ ways.
Hence the total number of seating arrangements = ⁵P₅ × ⁶P₃ = 5! × 6 × 5 × 4 = 14400.

Example 29. In how many ways can 9 examination papers be arranged so that the best and the worst papers are never together?
Answer:
The total number of arrangements of 9 papers = ⁹P₉ = 9!
Considering the best and the worst papers as one paper, we have 8 papers which can be arranged in ⁸P₈ = 8! ways. But the best and worst papers can be put together in 2! ways. So the number of permutations in which the best and the worst papers are together = 2! × 8!
Hence the number of ways in which the best and the worst papers never come together = 9! − 2! × 8! = 7 × 8! = 282240.

Example 30. How many four-digit numbers divisible by 4 can be made with the digits 1, 2, 3, 4, 5 if the repetition of digits is not allowed?
Answer:
Recall that a number is divisible by 4 if the number formed by its last two digits is divisible by 4.
With the digits 1, 2, 3, 4, 5 (no repetition), the tens and units digits can be: 12, 24, 32, 52, i.e. in 4 ways.
Corresponding to each such way, the remaining three digits can be arranged at the thousands and hundreds places in ³P₂ = 6 ways.
Hence the required number of numbers = ³P₂ × 4 = 6 × 4 = 24.

Example 31. Five boys and five girls form a line with the boys and girls alternating. Find the number of ways of making the line.
Answer:
5 boys can be arranged in a line in ⁵P₅ = 5! ways. Since the boys and girls are alternating, corresponding to each of the 5! arrangements of the 5 boys, we obtain 5 places for the girls, marked by crosses, in two possible patterns:
(i) B₁ × B₂ × B₃ × B₄ × B₅ ×
(ii) × B₁ × B₂ × B₃ × B₄ × B₅
Clearly, the 5 girls can be arranged in the 5 places marked by crosses in (5! + 5!) ways.
Hence, the total number of ways of making the line = 5! × (5! + 5!) = 2(5!)² = 28800.

Permutations of Objects Not All Distinct

Theorem: The number of mutually distinguishable permutations of n things, taken all at a time, of which p are alike of one kind and q are alike of a second kind such that p + q = n, is \( \frac{n!}{p!\,q!} \).
In general, if p things are alike of one kind, q alike of a second kind, r alike of a third kind and the rest are distinct, the number of permutations of all n things is \( \frac{n!}{p!\,q!\,r!} \).

Example 32. How many different words can be formed with the letters of the word MISSISSIPPI?
Answer:
There are 11 letters in the given word, of which 4 are S's, 4 are I's and 2 are P's.
So the total number of words is the number of arrangements of 11 things of which 4 are similar of one kind, 4 are similar of a second kind and 2 are similar of a third kind, i.e. \( \frac{11!}{4!\,4!\,2!} \)
Hence, the total number of words \( = \frac{11!}{4!\,4!\,2!} = 34650 \)

Example 33. How many different words can be formed by using all the letters of the word 'ALLAHABAD'? (i) In how many of them do the vowels occupy the even positions? (ii) In how many of them do both L's not come together?
Answer:
There are 9 letters in the word 'ALLAHABAD', out of which 4 are A's, 2 are L's and the rest are all distinct.
So the required number of words \( = \frac{9!}{4!\,2!} = 7560 \)
(i) There are 4 vowels and all are alike, i.e. 4 A's. Also there are 4 even places, viz. 2nd, 4th, 6th and 8th. So these 4 even places can be occupied by the 4 vowels in \( \frac{4!}{4!} = 1 \) way. Now we are left with 5 places, in which 5 letters, of which two are alike (2 L's) and the others distinct, can be arranged in \( \frac{5!}{2!} \) ways.
Hence the total number of words in which the vowels occupy the even places \( = \frac{5!}{2!} \times \frac{4!}{4!} = \frac{5!}{2!} = 60 \)
(ii) Considering both L's together and treating them as one letter, we have 8 letters out of which A repeats 4 times and the others are distinct. These 8 letters can be arranged in \( \frac{8!}{4!} \) ways. So the number of words in which both L's come together \( = \frac{8!}{4!} = 1680 \)
Hence, the number of words in which both L's do not come together = total number of words − number of words in which both L's come together = 7560 − 1680 = 5880.

Example 34. How many arrangements can be made with the letters of the word 'MATHEMATICS'? In how many of them are the vowels together?
Answer:
There are 11 letters in the word 'MATHEMATICS', of which two are M's, two are A's, two are T's and all others are distinct. So the required number of arrangements \( = \frac{11!}{2! \times 2! \times 2!} = 4989600 \)
There are 4 vowels, viz. A, E, A, I. Considering these four vowels as one letter, we have 8 letters (M, T, H, M, T, C, S and one letter obtained by combining all the vowels), out of which M occurs twice, T occurs twice and the rest are all different. These 8 letters can be arranged in \( \frac{8!}{2! \times 2!} \) ways.
But the four vowels (A, E, A, I) can be arranged among themselves in \( \frac{4!}{2!} \) ways.
Hence the total number of arrangements in which the vowels are always together \( = \frac{8!}{2! \times 2!} \times \frac{4!}{2!} = 10080 \times 12 = 120960 \)

Circular Permutations

Permutations of objects in a row are called linear permutations. If we arrange the objects along a closed curve, viz. a circle, the permutations are known as circular permutations. Every linear arrangement has a beginning and an end, but there is nothing like a beginning or an end in a circular permutation. Thus, in a circular permutation, we consider one object as fixed and the remaining objects are arranged as in the case of linear arrangements.
Theorem: The number of circular permutations of n distinct objects is (n − 1)!
Note 1: In the above theorem, anticlockwise and clockwise orders of arrangement are considered as distinct permutations.
Note 2: If anticlockwise and clockwise orders of arrangement are not distinct, e.g. arrangements of beads in a necklace, arrangements of flowers in a garland etc., then the number of circular permutations of n distinct items is \( \frac{1}{2}[(n - 1)!] \).

Example 35. In how many ways can 8 students be seated in (i) a line (ii) a circle?
Answer:
(i) The number of ways in which 8 students can be seated in a line = ⁸P₈ = 8! = 40320.
(ii) The number of ways in which 8 students can be seated in a circle = (8 − 1)! = 7! = 5040.

Example 36. (i) In how many ways can 5 persons be seated around a circular table? (ii) In how many of these arrangements will two particular persons be next to each other?
Answer:
(i) 5 persons can be seated around a circular table in (5 − 1)! = 4! = 24 ways.
(ii) Considering the two particular persons as one person, we have 4 persons in all. These 4 persons can be seated around a circular table in (4 − 1)! = 3! ways. But the 2 particular persons can be arranged among themselves in 2! ways.
Hence the total number of arrangements = 3! × 2! = 12.

Example 37. If 20 persons were invited for a party, in how many ways can they and the host be seated at a circular table? In how many of these ways will two particular persons be seated on either side of the host?
Answer:
(i) Clearly, there are 21 persons, including the host, to be seated round a circular table. These 21 persons can be seated round a circular table in (21 − 1)! = 20! ways.
(ii) Let P₁, P₂ be the two particular persons and H be the host. These two particular persons can be seated on either side of the host in the following two ways: P₁HP₂ and P₂HP₁.
Considering the two particular persons and the host as one person, we have 18 + 1 = 19 persons in all. These 19 persons can be seated round a circular table in (19 − 1)! = 18! ways.
So, the number of ways of seating the 21 persons at the circular table with the two particular persons on either side of the host = 18! × 2.

Example 38. Find the number of ways in which 8 different flowers can be strung to form a garland so that 4 particular flowers are never separated.
Answer:
Considering the 4 particular flowers as one flower, we have 5 flowers, which can be arranged in a circle in (5 − 1)! = 4! ways; since a garland looks the same from both sides, this gives \( \frac{1}{2} \times 4! \) different garlands.
But the 4 particular flowers can be arranged among themselves in 4! ways.
Hence the required number of ways \( = \frac{1}{2} \times 4! \times 4! = 288 \). (If the clockwise and anticlockwise arrangements are counted as different, the number is 4! × 4! = 576.)

Example 39. Find the number of ways in which 10 different beads can be arranged to form a necklace.
Answer:
Ten different beads can be arranged in circular form in (10 − 1)! = 9! ways.
Since there is no distinction between the clockwise and anticlockwise arrangements, the required number of arrangements \( = \frac{1}{2} \times 9! = 181440 \)