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(IGP) IAS Pre Paper - 2: GS - Basic Numeracy - Permutation & Combination
Basic Numeracy
Permutation & Combination
Fundamental Principles of Counting
Fundamental Principle of Multiplication: If there are two jobs such that one of them can be completed in n ways and second job can be completed in n ways, then the two jobs in succession can be completed in m × n ways. 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 its can be performed in (m + n) ways.
Example 1: Find the number of ways in which n different prizes can be distributed among m (< n) persons, if each is entitled to receive at most n – 1 prizes. Solution. Total number of ways = m × m × ... to n times = mn The number of ways in which one gets all the prizes = m The required number of ways = mn – m
Example 2: There are 4 candidates for the post of a lecturer in
Mathematics and one is to be selected by votes of 5 men. Find the number of ways
in which the votes can be given.
Solution. Each man can vote for one of the 4 candidates and this can be
done in 4 ways. Similar is the case with every other man. Hence, 5 men can
vote in 45 ie, 1024 ways.
Permutations
Each of the arrangements which can bemade by taking some or all number of things is called a permutation. Thus, the permutations which can be made by taking the letters a, b, c two at a time are 6. ie, ab, bc, ac, ba, cb and ca. Each of these presenting a different arrangement of two letters. These six arrangements are called permutations of three things taken two at a time.
Linear Permutation
If the things are arranged in a row/line, then a permutation is called linear permutation. It is simply written as a permutation.
- The number of permutations of n dissimilar things taken r at a time is
nPr.
nPr. = n(n – 1)(n – 2)…(n – r – 1) = n!/ ( n-r)! - Where n! is the product of the first n natural numbers and called ‘n –
factorial’ or ‘factorial n’ denoted by n! or n eg, 5! = 5 × 4 × 3 × 2 × 1 =
120 Here, we also define that 10 or 0 is 1. Also, n! is defined only for
positive integers. eg, The number of permutations of 6 dissimilar things
taken 4 at a time is
6P4 = = 6 × 5 × 4 × 3 = 360 - The number of permutations of n dissimilar things taken all at a time is
nPn .
nPn = n!/( n-n)!= n! - eg, The number of permutations of 3 dissimilar things taken all at a time is 3P3 = 3!/ (3-3)! = 3! = 3 × 2 × 1 = 6
- The number of permutations of n dissimilar things taken r at a time, when repetition of things is allowed any number of times is nr . eg, The number of different telephone numbers formed by taking 3 digits from 1, 2, 3, 4 is 43 = 64
- If n dissimilar things are grouped into k groups such that al things in
the first group are alike,… a2 in the second group are alike ak
in the kth group are alike, then the number of permutation of n such things
taken all at a time is
eg, The number of permutations that can be made using all the letters of the word ‘MANORAMA’ is 8/2.3 (Since, the given word contains 8 letters of which there are 2M’s, 3A’s and three different letters.)
Circular Permutation
If the things are arranged around a circle, then a permutation is called a circular permutation. In circular permutation, there is no first or last place of an object.Hence, the principles of linear permutations are not applicable in circular permutations. In such type of permutations, the relative positions of the things alone need to be taken into consideration and not the actual position.
- The number of circular permutation of n different things taken all at a time around a circle is (n - 1)! . eg, The number of ways in which 6 students sit around the table is (6-1)! ie, 5!. Circular Permutat ion Around a Thread
- If any arrangement of n different things around a circle in the clockwise direction is considered to be not different from the similar arrangement in the anti- lockwise direction, then, in this case, the number of circular permutations of n different things is 1/2(n-1)!.
- In the arrangement of flowers in a closed garland or beads in a
necklace, no distinction need to be made between clockwise and
anti-clockwise directions. eg, The number of ways in which 8 differently
coloured beads be strung on a necklace is
Combinations
Each of the groups or selections that can be made by taking some or all of a number of things is called a combination. Thus, the selections that can bemade by taking the letters a, b, c two at a time are three. ie, ab, bc and ac. These three groups or selections are called the combinations of three things taken two at a time.
- The number of combinations of n dissimilar things taken r at a time is nCr.
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The number of selections of 3 dissimilar things taken all at a time is 3C3=3!/(3-3)!= 1
Distribution of Things into Groups
- The number of ways in which things can be divided into two different
groups ofmand n things respectively
The number of ways of dividing 10 books into two groups each containing 7 and
3 books is
If m = n, then, we have to divide the given items into 2 equal groups, then 2 cases arise
- When the two groups have distinct identity As m= n, we get the number of
ways of dividing 2mthings into
- When the two groups do not have distinct identity. In this case, we have
to divide the above result by 2! as the two groups can be interchanged in
between themselves in 2! ways. Again as m = n,
[Here, no distinction is made between one parcel and other so the order is immaterial. The number of ways in which (m + n + r) things can be divided into three groups containing m, n and r
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Total Number of Combinat ions
- The total number of combinations of (p + q) things taken any number at a time when p things are alike of one kind and q things are alike of a second kind is (p + 1) q + 1) – 1. eg, There are 4 oranges, 5 pineapples and 7 watermelons in a fruit basket. The number of ways in which a person can choose one or more of the fruits is [(4 + 1) × (5 + 1) × (7 + 1)] – 1 = (5 × 6 × 8) –1 = 239
- The total number of ways of selecting one or more objects from n given objects is (2n – 1). eg, The number of ways in which a man can invite one or more of his 5 friends to dinner is (25 – 1) = 32 – 1= 31 ways.
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Example 5 An automobile dealer provides cars in 3models and 7
different colours each. Find the number of choices open to a customer.
Solution. Using the fundamental principle, number of choices open to a
customer = 7 × 3 = 21
Example 6: How many, permutations can be made
(i) Using all the letters of the word ‘COMPUTER’
(ii) How many begin with C ?
Solution. (i) Since, all the letters are distinct that are 8 in number.
The required number of permutations is 8P8 – 8!.
(ii) When C occupies the first
place the remaining 7 letters can be arranged among themselves in 7P7 = 7! ways.
Example 7: How many 4 letter permutations can be made from the letters of the
word FLOWER?
Solution. Since, all the 6 letters in the word are distinct, the number of
permutations
Example 8: Find the number of 3 digit numbers that can be formed from the digits (i) {1, 2, 3, 4, 5} (ii) {1, 2, 3, 4, 5, 6, 7, 8} when the digits are not repeated.
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Example 9: Find the number of 3 digit numbers that can be formed from the
digits (i) {1, 2, 3, 4, 5} (ii) {1, 2, 3, 4, 5, 6, 7, 8} using the digits any
number of times. Solution. (i) The numbers of 3 digit numbers formed by
using the given 5 numbers any number of times is 5 × 5 × 5 = 53.
(ii) The number of 3 digit numbers formed by using the given 8 numbers any
number of times is 8 × 8 × 8 = 83.
Example 10: Find the number of permutations that can be made using all
the letters of the word
(i) MISSISSIPI (ii) CHEESE.
Example 11: How many words can be formed using all the letters of the
word ‘MONDAY’ so that
(i) the vowels always come together
(ii) the vowels are never together?
Solution. (i) Given word contains 6 different letters when the vowels OA
are always together, we may suppose them to form an entity, treated as one
letter. Then, the letters to be arranged are MNDY (OA). These 5 letters
can be arranged in
5P5 = 5! = (5 × 4 × 3 × 2 × 1) ways = 120 ways.
The vowels in the group (OA) may be arranged in
2P2 = 2! = 2 × 1= 2 ways.
Required number of words = 120 × 2 = 240
(ii) The total number of words formed by using all the 6 letters of the given
word = 6P6 = 6! = (6 × 5 × 4 × 3 × 2 ×1) = 720
From (i) the numbers of words in which the vowels are always together = 240
:. The number of words in which the vowels are never together = (720-240) = 480
Example 12: In how many ways 10 persons can sit around around table?
Solution. The number of ways in which 10 persons can sit around around
table is (10 – 1)! = 9!.
Example 13: Find the number of ways in which 7 women can sit around a
round table so that all shall not have the same neighbours in any two
arrangements.
Solution. Circular arrangement of 7 women in one direction only (7-1)!/2
= 6!/2 = 720/ 2 = 360
Example 14: Find the number of ways in which 5 players out of 8
players can be selected to form a team is
Solution. The required number of ways is
Example 15: Find the number of ways in which a committee consisting of
3men and 2 women can be formed from 5 men and 4 women.
Solution. 3 men out of 5 can be selected in 5C3
ways. 2 women out of 4 can be selected in 4C2 ways. By the
fundamental principle, these two can be done in 1C3 ×
4C2 ways.
Example 16: Find the number of ways in which 6 players out of 10
players can be selected so as 2 particular players are always included.
Solution. Since, 2 particular players are always included the choice
reduces to only 4 players among 8 players which can be done in 8C4
ways.
Example 17: A set of m parallel lines intersect another set of n
parallel lines in a plane. Find the number of parallelograms formed.
Solution. For a parallelogram to be formed we should have two parallel
lines to intersect another two parallel lines. Two out of m parallel lines and
two out of another n parallel lines can be chosen in nC2 × nC2 ways. The number
of parallelograms formed is mC2 × nC2
.
Example 18: The number of ways in which the sixfaces of a cube be
painted with six different colours.
Solution. Number of ways = 6! / 6! = 1 (All faces are alike).
Example 19: There are stalls for 10 animals in a ship. The number of
ways the shipload can bemade if there are cows, calves and horses to be
transported, animals of each kind being not less than 10.
Solution. Each stall can be filed in 3 ways as there are three types of
animals (animals of each category being not less than 10). Shipload, ie, filling
up of 10 stalls,can be made in 3 × 3 × ... up to 10 times = 310.
Example 20: An examination paper contains 8 question of which 4 has 3
possible answers, 3 have 2 possible answers each and the remaining one question
has 5 possible answer. The total number of possible answers to all the questions
Solution. Total number of possible answers = 34 × 23
× 51 = 81 × 8 × 5 = 3240
Example 21: Manoj, Manika, Shivangi, Rohit are to give speeches in a
class. The teacher can arrange the order of their presentation in number of
ways.
Solution. Required number of ways = 4! = 24
Example 22: If there are 12 persons in a party, and if each of them
shakes hands with each other, then number of handshakes happen in the party.
Solution. Total number of handshakes = The number of ways of selecting 2
persons from among 12 persons = 12C2 = (12×11)/(2×1) = 66
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