Permutation and Combination Questions have always been popular among CAT aspirants.
In CAT, questions from permutation and combination are mostly based on the concepts of fundamental principles of counting. There is very rare occasions when the questions on P & C require a higher concept to solve.
We advice CAT aspirants to stick to the basics and solve permutation and combination questions which requre basic concepts and formulas.
They should not waste their time and energy to solve IIT JEE level questions unnecessarily.
For Practice and reference, we are providing few selected types of problems on permutation and combination which we are expected to appear in CAT exam.
Question 1:
A train is going from Mumbai to Pune and make 5 stops on the way. Three persons enter the train after it has started from Mumbai with 3 different tickets. How many different sets of tickets may they have had?
[1] \({}^{15}{P_3}\)
[2] \(3 \times {}^{15}{P_3}\)
[3] \({}^{15}{C_3}\)
[4] \(\frac{{{}^{15}{C_3}}}{3}\)
Answer & Solution
Option # 3
Since three persons enter the train after it has started from Mumbai, so they could have entered at:
First station (from where they could have bought tickets for the 2nd, 3rd, 4th or 5th station or for Pune i.e. 5 tickets).
Second station (from where they could have bought tickets for 3rd, 4th or 5th stations or for Pune i.e. 4 tickets)
3rd station (from where they could have bought tickets for the 4th or 5th stations or for Pune ie 3 tickets).
4th station (from where they could have bought tickets for the 5th station or for Pune i e. 2 tickets).
5th station (from where they could have bought a ticket for Pune i.e. 1 ticket)
Thus, we can see that there are 5+4+3+2+1=15 tickets available out of which 3 tickets are to be selected This can be done in \({}^{15}{C_3}\) ways.
Question 2:
The sum of all the possible numbers of 4 digits formed by digits 3, 5, 5, and 6 using each digit once is
[1] 64427
[2] 63327
[3] 65297
[4] 43521
Answer & Solution
Option # 2
Total number of different numbers of 4 digits formed by using these digits only once =\(\frac{{4!}}{{2!}} = 12\). The numbers formed whose unit digit is 3 = 3
The number formed whose unit digit is 5 = 6
The number formed whose unit digit is 6 = 3
(Same rule is applicable for 10th, 100th and 1000th units)
Sum of the total number of unit digits = 3×3+5×6 +6×3 = 57
Sum of the 10th digit number = 57×10
Sum of the 100th digit number = 57×100
Sum of the 1000th digit number = 57×1000
Sum of all the numbers formed = 57(1+10+100 + 1000) = 63327.
Question 3:
India and Australia player one-day international cricket series until anyone team win 4 matches. No match ended in a draw. In how many ways can the series be won?
[1] 35
[2] 70
[3] 105
[4] 140
Answer & Solution
Option # 2
No. of Matches | India Wins | Australia Wins | No. of Ways |
4 | 4 | \({}^4{C_4} = 1\) | |
5 | 4 | 1 | \({}^4{C_3} = \frac{{4!}}{{3!}} = 4\) |
6 | 4 | 2 | \({}^5{C_3} = \frac{{5!}}{{3! \times 2!}} = 10\) |
7 | 4 | 3 | \({}^6{C_3} = \frac{{6!}}{{3! \times 3!}} = 20\) |
So, India can win the series in 1+4+10+20 = 35 ways. Similarly, Australia can win the series in 35 ways. Hence, the total number of ways in which the series can be won is 35+35=70 ways.
Question 4:
Spider has one sock and one shoe for each of its 8 legs. In how many different ways can the spider put on its socks and shoes, assuming that in each leg the sock has to be put on before the shoe?
[1] \(\frac{{{{\left( {8!} \right)}^2}}}{2}\)
[2] \(\frac{{16!}}{{{2^8}}}\)
[3] \(\frac{{{{\left( {16!} \right)}^2}}}{{{2^8}}}\)
[4] None of These
Answer & Solution
Option # 2
Let the spider's legs be numbered 1. 2, 3, …, 8.
Let a_{k}, and b_{k}, denote the Sock and the shoe respectively for the k^{th}leg where 1 ≤ k ≤ 8.
So a total of sixteen objects are there (a_{1}b_{1}a_{2}b_{2}a_{3}b_{3}…. a_{8}b_{8}) which are to be arranged such that a_{k}comes before b_{k}
16 objects can be arranged in 16! ways, Out of these the arrangements in which a_{k}comes before b_{k=}\(\frac{1}{2} \times 16!\)
Similarly out of the remaining, the arrangement in which a_{2}comes before b_{2}
=\(\frac{1}{2} \times \frac{1}{2} \times 16!\) and so on
Finally arrangements in which a_{k}comes before b_{k}=\({\left( {\frac{1}{2}} \right)^8} \times 16!\)
Question 5:
Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition). the total number of such numbers that can be formed is
[1] 216
[2] 214
[3] 212
[4] 200
Answer & Solution
Option # 1
Sum of all given numbers = 0 + 1 + 2 + 3 + 4 + 7 + 8 = 25
Hence we will select only those five numbers whose sum is 18 or 9. But no five numbers will make the sum 9 So we choose the five numbers which make their sum 18 Since the sum of all the seven digits is 25 so we exclude those two digits out of given seven digits whose sum is 7 Thus we exclude either 0, 7 or 3, 4 as both these have sum 7. Hence we have the following two sets
[(a) 1, 2, 3, 4, 8, (0, 7 excluded) Therefore, sum = 18
They can be arranged in 5! = 120 ways
[(b) 1, 2, 7, 8, (3, 4 excluded) Therefore, sum = 18
They can be arranged in 5! - 4! (0, in beginning) = 120 - 24 = 96 ways T
Thus the total number of numbers divisible by 9 is 120 + 96 = 216
Question 6:
What is the number of whole numbers formed on the screen of a calculator which can be recognised as numbers with (unique) correct digits when they are read inverted? The greatest number that can be formed on the screen of the calculator is 999999.
[1] 100843
[2] 10843
[3] 1000843
[4] 108543
Answer & Solution
Option # 1
The digits which can be recognised as unique digits when they are Inverted in a calculator are 0, 1, 2, 5, 6, and 9 Since the number cannot begin with zero all the numbers having 0 at units place should be discarded for otherwise when reed upside down the number yen begin with 3 we now :1St the different posibililities
Number of Digits | Total number of numbers |
1 | 7 |
2 | $6 \times 6 = 6 ^ { 2 }$ |
3 | $6 \times 7 \times 6 = 6 ^ { 2 } \times 7$ |
4 | $6 \times 7 \times 7 \times 6 = 6 ^ { 2 } \times 7 ^ { 2 }$ |
5 | $6 \times 7 \times 7 \times 7 \times 6 = 6 ^ { 2 } \times 7 ^ { 3 }$ |
6 | $6 \times 7 \times 7 \times 7 \times 7 \times 6 = 6 ^ { 2 } \times 7 ^ { 4 }$ |
Thus, the number of required numbers $= 7 + 6 ^ { 2 } + 6 ^ { 2 } \times 7 + \ldots + 6 ^ { 2 } \times 7 ^ { 4 }$
$= 7 + 6 ^ { 2 } \frac { \left( 7 ^ { 3 } - 1 \right) } { 7 - 1 } = 7 + 6 \left( 7 ^ { 5 } - 1 \right) = 100843$ . Ans.(1)
Question 7:
In how many ways 2 different numbers can be chosen using the numbers between 0 and 180 (both inclusive) so that 60 is their average?
[1] 50
[2] 60
[3] 80
[4] None of these
Answer & Solution
Option # 2
Let x and y be 2 numbers such that $\frac { x + y } { 2 } = 60 \Rightarrow x + y = 120$ x and y both cant be equal to or greater than 60 (Since 60 cant be used twice)
let $0 \leq x \leq 59$ and $61 \leq y \leq 120$
The total number of ways in which x can be chosen = $60C _ { 1 } = 60$ ways and the value of y depends on the value of x and there will be only one value of y corresponding to a value of x.
Therefore, total number of ways in which the numbers can be formed = 60. Ans.(2)
Question 8:
The number of ways in which 4 squares can be chosen at random on a chess board such that they lie on a diagonal line?
[1] 140
[2] 182
[3] 364
[4] 256
Answer & Solution
Option # 3
Divide the chess board into two parts by a diagonal line 80 To select 4 squares lying on diagonal line we have to choose the squares lying on $P _ { 1 } , Q _ { 1 } , P _ { 2 } Q _ { 2 } , P _ { 3 } , Q _ { 3 } , P _ { 4 } Q _ { 4 }$ and $B D$.
Number of squares on one side of $B D = ^ { 4 } C _ { 4 } + ^ { 5 } C _ { 4 } + ^ { 6 } C _ { 4 } + 7 C _ { 4 } = 56$. Thus total squares on either side of BO = 56 x 2 = 112.
On the diagonal BD, we will have 8C4, squares=70. So total squares = 112 • 70 = 182.
Similarly on the other side of the diagonal line AC we will have 182 squares.
Hence total number of squares = 2 x 182 = 364 Ans.(3)
Question 9:
A Dubai based gangster Chhota Vakil is in Switzerland these jays. He wants to rob a bank there, whose locker code according to his information is an odd number between 50 and 450. He also knows that the numbers are from the set :0, 1, 2, 3, 4, 5}. How many maximum trials he has to take to unlock the locker?
[1] 54
[2] 72
[3] 78
[4] 106
Answer & Solution
Option # 2
Case (i): 2 digit number The tenth place digit can be 5 only (since number is greater than 50). The unit place can be filled in 3 ways. either 1 or 3 or 5 (as the number is odd number).
Case (ii): 3 digit number less than 400. 0 cannot be in the hundred's place. So the 100 hundred's place can be filled 3 ways (either 1 or 2 or 3) any digit can occupy the tenth place. Therefore it can be filled in 6 ways. The unit place can be filled in 3 ways. either 1 or 3 or 5. (as the number is odd number) Therefore number of ways = 3x 6x 3= 54.
Case (iii): number greater than 400 but less than 450. only 4 can come in the hundred's place. In the tenth place any digit can come except 5 (since number should be < 450), so it can be filled in 5 ways again the unit's place can be filled in three ways. number of ways = 1 x 5 x 3= 15.
Total number of ways = 3 + 54 + 15 = 72.
Question 10:
Given five different yellow balls, four different black balls and three different white balls. Then the number of combinations of balls that can be chosen taking atleast one yellow and atleast one black balls are
[1] 3255
[2] 4096
[3] 3720
[4] 3721
Answer & Solution
Option # 3
No. of ways in which at least one yellow ball can be selected = 25 - 1 = 31.
No. of ways in which at least one black ball can be selected = 2^{4}-1 = 15.
No. of ways in which one or more red balls can be selected = 2^{3} = 8,
Therefore. total number of ways = $\left( 2 ^ { 5 } - 1 \right) \left( 2 ^ { 4 } - 1 \right) 2 ^ { 3 } = 3720$
Question 11:
There are 4 letters and 4 addressed envelopes. If each letter is randomly placed in an envelope, then in how many ways can wrong choices be made?
[1] 4^{3}
[2] 4! – 1
[3] 16
[4] 4^{4}-1
Answer & Solution
Option # 2
4 letters can go into 4 envelopes in 4! Ways, of these onIy one choice is the correct one.
So. total number of wrong choices = 4! - 1 Ans.[2)
Question 12:
In the previous question, the number of ways in which only one letter goes in the wrong envelope is
[1] 43
[2] $^ { 4 } \mathrm { C } _ { 1 } + ^ { 4 } \mathrm { C } _ { 2 } + ^ { 4 } \mathrm { C } _ { 3 } + ^ { 4 } \mathrm { C } _ { 4 }$
[3] $^ { 4 } \mathrm { C } _ { 0 } + ^ { 4 } \mathrm { C } _ { 1 } + ^ { 4 } \mathrm { C } _ { 2 } + ^ { 4 } \mathrm { C } _ { 3 }$
[4] 0
Answer & Solution
Option # 4
At least two letters have to be interchanged for a wrong choice. So the number
of ways in which only one letter goes in wrong envelope = 0. Ans.(4)
Question 13:
In how many ways can the letters of the English alphabet be arranged so that there are seven letters between the letters A and B?
[1] 3! x 2!
[2] 24$\mathrm { C } _ { 7 } \times 18 ! \times 2$
[3] 36 x 24!
[4] 26$\mathrm { P } _ { 7 } \times 20 ! \times 2$
Answer & Solution
Option # 3
A and B can occupy the first and ninth places. the second and the tenth places, the third and eleventh place and so on upto eighteenth and twenty sixth place. This an be done in 15 ways. A and B can be arranged in 2 ways.
All the other 24 alphabets can be arranged m 24! ways. Hence. the required answer = $2 \times 18 \times 24 ! = 36 \times 24! .$ Ans.(3)
Question 14:
How many words can be formed with the letters of the word 'PATALIPUTRA' without changing the relative order of the vowels and consonants?
[1] 3600
[2] 4200
[3] 54000
[4] 3000
Answer & Solution
Option # 1
There are 11 letters in the word 'PATALIPUTRA‘ and there are two P's. two T's. three A's and four other different letters. Number of consonants = 6. number of vowels = 5.
Since relative order of the vowels and consonants remains unchanged. vowels can occupy only vowel’s place and consonants can occupy only consonants place.
Now 6 consonants can be arranged among themselves $\frac { 6! } { 2!2 ! }$ ways (since
there are two P's and two T’s).
And 5 vowels can be arranged among themselves in 5!/3!ways (since A occurs thrice)
Therefore, Required number = $\frac { 6!} { 2!2 ! } \times \frac { 5 ! } { 3 ! } = 3600$. Ans.(1)
Question 15:
A shop owner has an unlimited supply of 8 different types of flowers. If a customer asks for a five flower arrangement such that the arrangement should start and end with the same type of flower, then in how many ways the arrangement can be made by the shop owner?
[1] 8$C _ { 4 } \times 3 !$
[2] 8^{3}
[3] 8^{4}
[4] 8$C _ { 4 } \times 4 C _ { 1 } \times 3 !$
Answer & Solution
Option # 3
Chose with one flower which arrangement will start and end.
$8C1x8 ^ { 3 } = 8 ^ { 4 }$
Question 16:
There are 21 balls which are either white or black and the balls of the same colour are alike. If the number of arrangements of these balls in a row be maximum then the number of white balls can be
[1] 10
[2] 12
[3] 14
[4] 21
Answer & Solution
Option # 1
Let there be r white (alike) balls so that the number of black balls is (21-r)
(These are also alike). Number of arrangements of these total 21 balls is
$A=\frac{21!}{\text{ (r}!\times 21-r)!}{{=}^{21}}{{C}_{r}}$
Since r are alike of one kind and (21 - r) are alike of other kind, we have divided by r! and (21 - r)!
Again $A = ^ { 21 } C _ { r }$ , will be maximum when $r = \frac { 21 + 1 } { 2 }$ or $\frac { 21 - 1 } { 2 }$ i. e. 10 or 11
Question 17:
How many five letter words (with or without meaning) can be formed such that the letters appearing in the odd positions are taken from the unrepeated letters of the word MATHEMATICS whereas the letters which occupy even places are taken from amongst the repeated letters of the same?
[1] 3600
[2] 540
[3] -150
[4] None of these
Answer & Solution
Option # 2
There are 3 odd places namely lst, 3rd and 5th which are to be filled by unrepeated 5 letters H. E. l, C. S. This can be done in 5P3 = 5 X 4 X 3 = 60 ways. We have two even places namely 2nd and 4th which is to be filled by repeated letters (2M. 2A, 2T) i.e., 6 letters. These two even places can be filled by 3 different types of letters as under.
(i) All different: $: 3 P _ { 2 } = 3 \times 2 = 6$ .
(ii) Both alike $\therefore ^ { 3 } \mathrm { C } _ { 1 } \times \frac { 2 ! } { 2 ! } = 3$
Thus even places can be filled in 6 + 3 = 9 ways. Hence by fundamental principle of counting. the required number of words is 60 x 9 = 540. Ans.(2)
Question 18:
An entomologist noticed 15 ants crawling on his table. Exactly 13 of them were male and 2 were female. He noticed that they were crawling on the table in a random order such that no three of them were ever in the same straight line. Suddenly, three of the male ants start following one of the female ants such that, all four of them were in a single line. Then the maximum number of distinct straight lines that the entomologist can draw passing through any two ants is
[1] 56
[2] 44
[3] 105
[4] 100
Answer & Solution
Option # 4
Initially there were 15C2 = 105 lines. Now for the four ants among themselves the number of lines are decreased by (4C2 - 1) = 5 i.e., a net decrease of 5 lines i:e., there are 105 - 5 = 100 possible lines. Ans.(4)
Question 19:
A train is going from Mumbai to Pune and makes S stops on the way. 3 persons enter the train after it has started from Mumbai with 3 different tickets. How many different sets of tickets they may have had?
[1] 15$p _ { 3 }$
[2] $3 \times 15 P_ { 3 }$
[3] $15C _ { 3 }$
[4] $\frac { 15 C _ { 3 } } { 3 }$
Answer & Solution
Option # 3
Since three persons entered the train after it has started from Mumbai, so they could have entered at:
1st station (from where they could have bought tickets for the2nd, 3rd, 4^{th} or 5th stations or for Pune i.e. 5 tickets).
2nd station (from where they could have bought tickets for the 3rd, 4th for 5th stations or for Pune i.e. 4 tickets).
3rd station (form when: they could have bought tickets tor the 4th or 5th stations or tor Pune i.e. 3 tickets).
4th station from where they could have bought tickets tor the 5th station or for Pune i.e., 2 tickets)
5th station (from when: they could have bought a ticket for Pune i.e. 1 ticket)
Thus, we can see that there are 5+4+3+2+1= 15 tickets available out of which 3 tickets are to be selected. This can be done in 15C3 ways.
Question 20:
Suppose a city has m parallel roads running East-West and n parallel roads running North-South. How many rectangles are formed with their sides along these roads?
[1] $\frac { m n } { 4 }$
[2] $\frac { m n ( m - 1 ) ( n - 1 ) } { 4 }$
[3] $\frac { ( m - 1 ) ( n - 1 ) } { 4 }$
[4] None of these
Answer & Solution
Option # 2
To form a rectangle. we require two lines from East to West and two lines from North to South. Therefore, total number at rectangles which we can get is:
$mC _ { 2 } \times n C _ { 2 } = \frac { m ( m - 1 ) } { 2 } \cdot \frac { n ( n - 1 ) } { 2 } = \frac { m n ( m - 1 ) ( n - 1 ) } { 4 }$
Question 21:
Mohan, a thief, went to a liquor shop where he decided to take away 15 bottles. In the shop there are bottles of Wine, Whisky, Rum, Vodka and Gin. In how many ways he can select the bottles?
[1] 19C4
[2] 20C4
[3] 15C5
[4] 2200
Answer & Solution
Option # 1
As there is no restriction in choosing the bottles,
Therefore, By using the formula $^{n+r-1}{ C } _ { \mathrm { r } - 1 }$
$= ^{15 + 5 - 1}C _ { 5 - 1 } = ^ { 19 } C _ { 4 }$
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