TOPIC:

Permutation and

Combination

Permutation: Permutation means arrangement of things. The word arrangement is used, if the order of things is considered.

Combination: Combination means selection of things. The word selection is used, when the order of things has no importance.

Example:     Suppose we have to form a number consisting of three digits using the digits 1,2,3,4. To form this number the digits have to be arranged. Different numbers will get formed depending upon the order in which we arrange the digits.

This is an example of Permutation.

Now suppose that we have to make a team of 11 players out of 20 players.

This is an example of combination, because the order of players in the team will not result in a change in the team. No matter in which order we list out the players the team will remain the same! For a different team to be formed at least one player will have to be changed.

Addition rule : If an experiment can be performed in ‘n’ ways, & another experiment can be performed in ‘m’ ways then either of the two experiments can be performed in (m+n) ways. This rule can be extended to any finite number of experiments.

Two Fundamental Principles of Counting

Example:       Suppose there are 3 doors in a room, 2 on one side and 1 on other side. A man want to go out from the room. Obviously he has ‘3’ options for it. He can come out by door ‘A’ or door ‘B’ or door ’C’.

Multiplication Rule : If a work can be done in m ways, another work can be done in ‘n’ ways, then both of the operations can be performed in m x n ways. It can be extended to any finite number of operations.

Example.:      Suppose a man wants to cross-out a room, which has 2 doors on one side and 1 door on other site. He has  2 x 1  = 2 ways for it.

Factorial n : The product of first ‘n’ natural numbers is denoted by n!.

    n!   = n(n-1) (n-2) ………………..3.2.1.   Ex.       5! = 5 x 4 x 3 x 2 x 1 =120

 Note       0!     =  1

Permutation Number of permutations of ‘n’ different

things taken ‘r’ at a time is given by:-

nPr       =            n!/(n-r)! Note : Factorial of negative-number is not

defined. The expression  –3! has no meaning.

Examples Q. How many different signals can be made

by 5 flags from 8-flags of different colours?

Ans.    Number of ways taking 5 flags out of 8-

flags  = 8P5

=   8!/(8-5)!        =  8 x 7 x 6 x 5 x 4 = 6720

Q. How many words can be made by using the letters of the word “SIMPLETON” taken all at a time?

Ans.   There are ‘9’ different letters of the word “SIMPLETON”

Number of Permutations taking all the letters at a time = 9P9

=  9!    = 362880.

Number of permutations of n-thing, taken all at a time, in which ‘P’ are of one type, ‘g’ of them are of second-type, ‘r’ of them are of third-type, and rest are all different is given by :-

  n!/p!xq!xr!                                                                                               

Example: In how many ways can the letters of the word “Pre-University” be arranged?

Answer: 13!/2! X 2! X 2!

Number of permutations of n-things, taken ‘r’ at a time when each thing can be repeated r-times is given by = nr.

Example:      A child has 3 pocket and 4 coins. In how many ways can he put the coins in his pocket.

Answer First coin can be put in 3 ways, similarly

second, third and forth coins also can be put in 3 ways.

So total number of ways = 3 x 3 x 3 x 3   = 34   = 81

Combination Number of Combination of ‘n’ different

things, taken ‘r’ at a time is given by:-

nCr=  n! / r ! x (n-r)!                                 Note: nCr  =  nCn-r

Question 1: A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A. 1260 B. 210 C. 10C6 * 6! D. 10C5 * 6

Correct Answer - 1260. Choice (a) Explanation

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.Now, the captain can be selected from these 6 players in 6 ways.Therefore, total ways the selection can be made is 210*6 = 1260.

Question 2:  How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?

a.16C7 * 7!b. 12C4 * 4C3 * 7!c. 12C3 * 4C4

d. 12C4 * 4C3

Correct Answer - (b) Solution:

4 consonants out of 12 can be selected in 12C4 ways.3 vowels out of 4 can be selected in 4C3 ways.Therefore, total number of groups each containing 4 consonants and 3 vowels

= 12C4 * 4C3

Each group contains 7 letters, which can be arranged in 7! ways.Therefore required number of words =

12C4 * 4C3 * 7!

Question 3:  If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?

A. 24 B. 31 C. 32 D. 30

Correct Answer - (c) Solution:

The 5 letter word can be rearranged in 5! Ways = 120 without any of the letters repeating.The first 24 of these words will start with A.Then the 25th word will start will CA _ _ _. The remaining 3 letters can be rearranged in 3! Ways = 6. i.e. 6 words exist that start with CA.The next word starts with CH and then A, i.e., CHA _ _. The first of the words will be CHAMS. The next word will be CHASM.Therefore, the rank of CHASM will be 24 + 6 + 2 = 32.

Question 4:  How many four letter distinct words can be formed using the alphabets of English language such that the last of the four letter word is always a consonant?

A. (263)*(21) B. 26*25*24*21 C. 25*24*23*21 D. None of these.

Correct Answer - (a) Solution:

The last of the four letter words should be a consonant. Therefore, there are 21 options.The first three letters can be either consonants or vowels. So, each of them have 26 options. Hence answer = 26*26*26*21 = 263 * 21

Question 5:  When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

155 620 671 625

Correct Answer - (c) Solution:

When 4 dice are rolled simultaneously, there will be a total of 64 = 1296 outcomes.The number of outcomes in which none of the 4 dice show 3 will be 54 = 625 outcomes.Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

Question 6:  A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

A. 9 B. 26 C. 126 D. 3920

Correct Answer - (c) Solution:

There are 8 students and the maximum capacity of the cars together is 9.We may divide the 8 students as followsCase I: 5 students in the first car and 3 in the second Or Case II: 4 students in the first car and 4 in the secondHence, in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.Therefore, the total number of ways in which 8 students can travel is 8C3 + 8C4 = 56 + 70 = 126.

Question 7:  There are 12 yes or no questions. How many ways can these be answered? a. 1024b. 2048c. 4096d 144

Correct Answer - (c) Solution:

Each of the questions can be answered in 2 ways (yes or no)Therefore, no. of ways of answering 12 questions = 212 = 4096 ways.

Question 8:  How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively?

A. 5! B. 4! C. 6! - 2! D. 6! / 2!

Correct Answer - (b) Solution:

As A and T should occupy the first and last position, the first and last position can be filled in only one way. The remaining 4 positions can be filled in 4! ways by the remaining letters (S,C,E,N). hence by rearranging the letters of the word ASCENT we can form 1x4! = 4! words.

Question 9:  Four dice are rolled simultaneously. What is the number of possible outcomes in which at least one of the die shows 6?

a. 6! / 4! b. 625 c. 671 d. 1296

Correct Answer - (c) Solution:

When 4 dice are rolled simultaneously, there are 64 = 1296 outcomes. The converse of what is asked in the question is that none of the dice show '6'. That is all four dice show any of the other 5 numbers. That is possible in 54 = 625 outcomes.

Therefore, in 1296 - 625 = 671 outcomes at least one of the dice will show 6.

Question 10:  In how many ways can the letters of the word MANAGEMENT be rearranged so that the two As do not appear together?

A. 10! - 2! B. 9! - 2! C. 10! - 9! D. None of these

Correct Answer - (4) Solution:

The word MANAGEMENT is a 10 letter word. Normally, any 10 letter word can be rearranged in 10! ways. However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters A, M, E and N repeat twice each. Therefore, the number of ways in which the letters of the word MANAGEMENT can be rearranged reduces to

The problem requires us to find out the number of outcomes in which the two As do not appear together.The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter. Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in ways.

Therefore, the required answer in which the two As do not appear next to each other =Total number of outcomes - the number of outcomes in which the 2 As appear together=> ways.

Question 11:  There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs. How many different albums can be formed using the above repertoire if the albums should contain at least 1 Rock song and 1 Carnatic song?

A. 15624 B. 16384 C. 6144 D. 240

Correct Answer - (a) Solution:

There are 2n ways of choosing ‘n’ objects. For e.g. if n = 3, then the three objects can be chosen in the following 23 ways - 3C0 ways of choosing none of the three, 3C1 ways of choosing one out of the three, 3C2 ways of choosing two out of the three and 3C3 ways of choosing all three.In the given problem, there are 5 Rock songs. We can choose them in 25 ways. However, as the problem states that the case where you do not choose a Rock song does not exist (at least one rock song has to be selected), it can be done in 25 - 1 = 32 - 1 = 31 ways.Similarly, the 6 Carnatic songs, choosing at least one, can be selected in 26 - 1 = 64 - 1 = 63 ways.And the 3 Indi pop can be selected in 23 = 8 ways. Here the option of not selecting even one Indi Pop is allowed.Therefore, the total number of combinations = 31 * 63 * 8 = 15624

Question 12:  How many number of times will the digit '7' be written when listing the integers from 1 to 1000?

(1) 271(2) 300(3) 252(4) 304

Correct Answer - (2) Solution:

7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9. 1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etcThis means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7). There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 * 9 = 27 such numbers. In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times. 3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.Therefore, the total number of times the digit 7 is written between 1 and 999 is 243 + 54 + 3 = 300