PERMUTATION AND COMBINATION
Permutations are for the lists (order matters) and combinations are for groups (order doesn't matter). Combinations sound simpler than permutations. In this chapter, topics are related to factorial notation, combinations, permutations, fundamental theorem, and rank. Factorial Notation Let n be a positive integer. Then, factorial n, denoted n! is defined as n!=n(n - 1)(n - 2)....3.2.1
0! = 1
1!=1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
Permutations The different arrangements of a given number of things by taking some or all at a time, are called permutations.
Examples All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
All permutations made with the letters a, b, c taking all at a time are: (abc, acb, bac, bca, cab, cba)
A number of Permutation Number of all permutations of n things, taken r at a time, is given by: nPr = n(n-1)(n-2)....(n-r+1)= n!/(n-r)!
Each of the different group or selective which can be formed by taking some or all of a number of objects is called a combination.
Examples
Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC, and CA.
Note: AB and BA represent the same selection. A;; the combination formed by a, b, c taking ab, bc, ca. The only combination that can be formed of letters a, b, c taken all at a time is ABC. Various groups of 2 out of four-person A, B, C, D are: AB, AC, AD, BC, BD, CD.
Note that ab ba are two different permutations but they represent the same combination.
Number of Combinations The number of all combinations of n things, taken r at a time is:
nCr = n!/(r!)(n-r)!= n(n-1) (n-2)....to r factors/r! Note nCn = 1 and nC0= 1. nCr = nC(n-r)

SIMPLE PERMUTATION

Question 1 In a class, there are 10 boys and 8 girls. The class teacher wants to select a boy and a girl torepresent the class in a function. In how many ways can the teacher make this selection?
Solution
The teacher has to perform 2 jobs:
To select a boy among 10 boys, which can be done in 10 ways.
To select a girl, among 8 girls, which can be done in 8 ways.
Hence, the required number of ways= 10*8=80

CIRCULAR PERMUTATION

(a)Arrangements at a Circular Table
Question 2 In how many different ways can 5 boys and 5 girls form a circle such that boys and girls are alternate?
Solution
After fixing up 1 boy at the table the remaining can be arranged in 4! Ways.
There will be 5 places, one place each between two boys which can be filled by 5 girls in 5! Ways.
Hence by the principle of Multiplication, the required number of ways= 4!* 5!= 2880.
(b)Arrangements of beads of flowers (all different) around a Circular Necklace or Garland

Question 3 Find the number of ways in which 18 different beads can be arranged to form a necklace?
Solution
18 different beads can be arranged among themselves in a circular order in (18-1)!= 17! Ways
Now in the case of necklace, there is no distinction between clockwise and anticlockwise arrangements.
So the required number of arrangements = ½ (17!) = 17!/2

FORMATION PERMUTATION

(a)Word Forming
Question 4 How many different words can be formed with the letter of the word MISSISSIPPI?
Solution
In the word MISSISSIPPI, there are 41's, 4S's, and 2P's.
Thus required number of words=(11!) / 4! 2! 4! = 34650
Since there are 11 letters in all, therefore → 11!
(b)Number forming
Question 5 How many 4 digits numbers (Repetition is not allowed) can be made by using digits 1-7. If 4 will always be there in the number?
Solution
Total digits (n)= 7
Total ways of making the number if 4 is always there
=r x
n-1 P r-1
= 4 * 6P3
= 480

RANKING OF WORDS IN DICTIONARY

Question 6 If all permutations of the letter of the word AGAIN are arranged as in Dictionary, then the Fiftieth word is?
Solution
First, let's arrange all the letters in an Alphabetical order i.e AAGIN. Now, let’s start with the letter A,
and arrange in the other four letters, There are 4! = words. These are the first 24 words.
Then starting with G, and arranging A, A, I, and N in different ways, there are 4! / 2! 1! 1! = 24 / 2 = 12 words.
Hence, total of 36 words.
Next, the 37th word starts with I. there are 12 words starting with I. This accounts up to the 48th word. The 49th word is NAAGI. The 50th word is NAAIG.
A _ _ _ _= 4! =4*3*2*1= 24 Words
G _ _ _ _ = 4!/2!= 12 words
I _ _ _ _ = 4!/2!= 12 words
Till now total = 48 Words
N A A G I = 1 Word
N A A I G = 1 Word
Therefore the 50th word is NAAIG

CONDITIONAL COMBINATIONS

Question 8 In a class of 25 students, find the total number of ways to select two representatives.
If a particular person is never selected.
If a particular person is always there
Solution
(i) Total students (n) = 25
Particular students will not be selected (p) = 1
So the total number of ways 25-1C2 = 276
(ii) Using n-p Cr-p no. of ways = 25-1C2-1
24C₁ = 24
NOTE
If a person is always there, then we have to select only 1 from the remaining 25-1=24.

RECTANGLES AND SQUARES

Question 9 The number of squares that can be formed on a chess board is –
Solution.
A chessboard is made up of 9 equispaced horizontal and vertical line. To make 1 x 1 square, We must choose two consecutive horizontal and vertical lines from among these. This can be done in 8 x 8= 82 ways. A 2 x 2 square needs three consecutive horizontal and vertical lines and we can do this in 7 x 7= 72 ways. Continuing in this manner the total number of square is 8²+7²+6²+.....2² +1² = 204.
REPETITION
Allowed
Not allowed
Question 10 The number of 6-digit numbers that can be formed from the digits 1, 2, 3, 4.5 6, and 7 so that digits do not repeat do repeat?
Solution
(i) Do not repeat
The first digit can be any number from 1-7 = 7 ways
The second digit can be any number b/w 1-7 excluding the digit already taken above.= 6 ways
Similarly, 5 ways , 4 ways, 3 ways, 2 ways (since its we only need to form a 6 digit number)
7 * 6 * 5 * 4 * 3 * 2 = 5040 ways
(ii) Do Repeat
The first digit can be any number from 1-7 = 7 ways
Since repetition is allowed, then we can again use the number used above. Therefore there are
again going to be 7 ways
Similarly, since we only need to form a 6-digit number
Therefore, 76 = 117649 ways.