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permutation and combination problems with solutions



permutation and combination problems with solutions


Question1 : How many different 4-letter words can be formed using the letters of the word CONSTITUTION such it starts with C and ends with N?
Option 1 : 30
Option 2 : 33
Option 3 : 40
Option 4 : 24
Answer : B
Explanation : From the word CONSTITUTION, the 4-letter word
C_ _ N has to be formed.
Leaving out these two letters, we have
2 O, 1 N, 1 S, 3 T, 2 I, 1 U

3 unique letters - N S U
3 duplicate letters - O T I

If our second letter is either of N,S or U
we have 5 choices for the third letter.
Eg: NS, NU, NO, NT, NI

If our second letter is either O, T or I
we have 6 choices for the third letter.
Eg: ON, OS, OU, OO, OT, OI

No. of ways
= (3*5)+(3*6)
= 33

Question2 : What is the value of 1*1! + 2*2! + 3!*3! + ............ n*n! ?
Option 1 :  n(n-1)(n-1)!
Option 2 : (n+1)!/(n(n-1))
Option 3 : (n+1)! - n!
Option 4 : (n + 1)! - 1!
Answer : D
Explanation : 1*1! = (2 -1)*1! = 2*1! - 1*1! = 2! - 1!

2*2! = (3-1)*2! = 3*2! - 1*2! = 3! - 2!

n*n! = (n+1 - 1)*n! = (n+1)(n!) - n! = (n+1)! - n!

1*1! + 2*2! +.......... n*n!
= 2! - 1! + 3! - 2! +.....+(n+1)! - n!
= (n+1)! - 1!

Question 3: How many 3 letter words, with or without meaning, can be formed using the letters of the English alphabet without repetition such that the word begins with a vowel and ends with a consonant?
Option 1 : 50
Option 2 : 120
Option 3 : 504
Option 4 : 2520
Answer : D
Explanation : The first letter is one of the 5 vowels.
No. of ways
= 5

The last letter is one of the 21 consonants.
No. of ways
= 21

The middle letter is selected from the remaining 24 letters.
No. of ways
= 24

Total no. of ways
= 5*21*24
= 2520

Question4 : From a group of 6 men and 8 women, six persons are to be selected to form a committee such that there are at least 2 men and 2 women in the committee. In how many ways can it be done?
Option 1 : 300
Option 2 : 1120
Option 3 : 1050
Option 4 : 2590
Answer : D
Explanation : The commitee can consist of
2 men, 4 women
3 men, 3 women
4 men, 2 women
No. of ways
= (6C2*8C4)+(6C3*8C3)+(6C4*8C2)
= (15*70)+(20*56)+(15*28)
= 1050+1120+420
= 2590

Question5 : In how many ways can 5 letters be posted in 4 postboxes if each postbox can contain any number of letters?
Option 1 : 125
Option 2 : 625
Option 3 : 1024
Option 4 : 20
Answer : C
Explanation : No. of ways of posting the first letter in any of the 4 postboxes
= 4
Similarly, no. of ways of posting the second letter
= 4
No. of ways of posting the 5 letters
= 4*4*4*4*4
= 1024

Question6 : Find the number of triangles in a heptagon.
Option 1 : 28
Option 2 : 35
Option 3 : 21
Option 4 : 14
Answer : B
Explanation : There are 7 vertices in a heptagon.

No. of ways of selecting 3 vertices out of 7
= 7C3
= 35

Question7 : A committee consisting of 3 men and 4 women is to be formed from a group of 7 men and 8 women. In how many ways can this be done?
Option 1 : 105
Option 2 : 2450
Option 3 : 700
Option 4 : 350
Answer : B
Explanation : No. of ways of selecting 3 men from 7
= 7C3
= 35
No. of ways of selecting 4 men from 8
= 8C4
= 70
Total no. of ways
= 35*70
= 2450

Question8 : In how many ways can four cards of different suits having the same value be selected from a standard pack of cards?
Option 1 : 13
Option 2 : 52
Option 3 : 26
Option 4 : 169
Answer : A
Explanation : A card can take 13 distinct values.

Since all 4 cards have the same value, only the value has to be selected from the 13 values.

No. of ways of selecting 1 value from 13 values
= 13C1
= 13

Question9 : In how many ways can you select four cards, each of a different suit from a standard deck of cards?
Option 1 : 208
Option 2 : 520
Option 3 : 2197
Option 4 : 28561
Answer : D
Explanation : No. of ways of selecting a spade from 13 spades
= 13C1 = 13
No. of ways of selecting a clubs from 13 clubs
= 13
No. of ways of selecting a hearts from 13 hearts
= 13
No. of ways of selecting a diamond from 13 diamonds
= 13
Total no. of ways
= 13*13*13*13
= 28561

Question10 : Find the number of possible ways in which 8 students can be divided into 4 equal teams to take part in an inter-school competition.
Option 1 : 2520
Option 2 : 1260
Option 3 : 32
Option 4 : 49
Answer : A
Explanation : Each team consists of 2 students.

For the first team,
2 students are selected from 8 in
8C2 = 8*7/2 = 28 ways
For the second team,
2 are selected from the remaining 6 in
6C2 = 6*5/2 = 15 ways
For the third team,
2 are selected from the remaining 4 in
4C2 = 6 ways
The remaining 2 will belong to the fourth team.

Total no. of ways
= 28*15*6
= 2520

Question11 : In how many ways can the letters of the word EDUCATE be rearranged so that the two Es do not appear together?
Option 1 : 960
Option 2 : 1800
Option 3 : 720
Option 4 : 2520
Answer : B
Explanation : No. of ways of arrangin the letters of EDUCATE (7 letters, 2 Es)
= 7!/2!
= 2520

If EE is taken as a single letter, there would be a total of 6 letters.

No. of ways in which E's appear together
= 6!
= 720
Required no. of ways
= 2520-720
= 1800

Question12 : How many 3-letter words with or without meaning, can be formed out of the letters of the word SUCTION if repetition of letters is not allowed?
Option 1 : 210
Option 2 : 840
Option 3 : 21
Option 4 : 140
Answer : A
Explanation : SUCTION contains 7 different letters.

No. of ways of arranging 3 letters out of 7 letters
= 7P3
= 7!/4!
= 210

Question13 : A family of 10 people travel in two cars, of which one can seat 6 and the other only 5. In how many ways can they travel?
Option 1 : 252
Option 2 : 500
Option 3 : 462
Option 4 : 450
Answer : C
Explanation : They can travel in these ways.
6 people in the first car, 4 people in the second car
5 people in the two cars
Required no. of ways
= 10C6+10C5
= 210+252
= 462

Question14 : 8 friends go for a trip in 2 cars. Each car can accomodate a maximum of 4 persons. In how many ways can they travel?
Option 1 : 160
Option 2 : 140
Option 3 : 168
Option 4 : 70
Answer : D
Explanation : Required no. of ways
= No. of ways of selecting 4 persons for the first car (since the remaining four will have to travel in the second car)
= 8C4
= 8!/(4!.4!)
= 70

Question15 : In how many ways can 5 true or false questions be answered incorrectly?
Option 1 : 10
Option 2 : 25
Option 3 : 31
Option 4 : 32
Answer : C
Explanation : 5 true or false questions can be answered in 2^5=32 ways.

Only one of these is correct.

No. of ways of answering incorrectly
= 31

Question16 : Find the number of ways in which 11 players can be selected from 15 players if the captain and the vice-captain are always selected.
Option 1 : 650
Option 2 : 715
Option 3 : 165
Option 4 : 117
Answer : B
Explanation : The captain and the vice-captain are always selected.

The rest of the 9 players are to be selected from the remaining 13.

No. of ways
= 13C9
= 13!/(9!*4!)
= 715

Question17 : There are 8 seats in a railway compartment - 4 seats facing the engine and the other 4 facing away from the engine. Of the 8 passengers, 3 prefer to face the engine, 3 prefer away and 2 have no preference. In how many ways can the passengers be arranged based on their preferences?
Option 1 : 1152
Option 2 : 576
Option 3 : 24
Option 4 : 2
Answer : A
Explanation : Of the 4 seats facing the engine,
3 are occupied by the persons who prefer it.
No. of ways of arranging 3 persons in 4 seats = 4!
Of the 4 seats facing away from the engine,
3 are occupied by the persons who prefer it.
No. of ways of arranging 4 persons in 3 seats= 4!
The rest 2 are occupied by the persons who have no preferences. It can be done in 2 ways.

Required no. of ways
= 2*4!*4!
= 1152

Question18 : Find the number of diagonals in an octagon.
Option 1 : 8
Option 2 : 16
Option 3 : 28
Option 4 : 20
Answer : D
Explanation : An octagon has 8 vertices.
No. of ways of connecting 2 vertices
= 8C2
= 8*7/2
= 28
Subtracting the 8 sides,
the no. of diagonals = 28-8 = 20

Question19 : In a party, there are 50 people. Each person shakes their hand with every other person. How many handshakes take place?
Option 1 : 1225
Option 2 : 2450
Option 3 : 100
Option 4 : 500
Answer : A
Explanation : No. of handshakes
= No. of ways of selecting 2 people out of 50
= 50C2
= 50*49/2
= 1225

Question20 : How many 4-digit numbers can be formed from the digits 2, 4, 5, 6, 8 and 9 which are divisible by 5 and none of the digits is repeated?
Option 1 : 36
Option 2 : 25
Option 3 : 60
Option 4 : 120
Answer : C
Explanation : For the number to be divisible by 5, the last digit should be 5.

The first 3 digits can be one among 2,4,6,8 and 9.

Number of ways of arranging the first 3 digits
= 5P3
= 5!/2!
= 3*4*5
= 60

Question21 : In how many different ways can the letters of the word DENIAL be arranged in such a way that the vowels occupy only the even positions?
Option 1 : 720
Option 2 : 240
Option 3 : 36
Option 4 : 24
Answer : C
Explanation : The vowels E,I,A occupy only the even positions.

No. of ways of arranging the 3 vowels in the even positions
= 3!
No of ways of arranging the consonants in the odd positions
= 3!
Required no. of ways
= 3!*3!
= 36

Question22 : You have displayed your five favourite novels between the two ends of your book shelf. If you decide to arrange the five books in every possible combination and move just one book every minute, how long would it take you?
Option 1 : 24 minutes
Option 2 : 120 minutes
Option 3 : 10 minutes
Option 4 : 5 minutes
Answer : B
Explanation : No. of ways of arranging 5 books
= 5!
= 120
Time taken to arrange
= 120 minutes

Question23 : A school has 10 volleyball players. A team of 5 members and a captain has to be selected out of these 10 players. In how many different ways can the selection be done?
Option 1 : 210
Option 2 : 6!
Option 3 : 10C6 * 6!
Option 4 : 5040
Answer : A
Explanation : Out of the 10 players, 6 have to be selected.

Required no. of ways
= 10C6
= 210

Question24 : A box contains 2 blue balls, 3 green balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one red ball is to be drawn?
Option 1 : 74
Option 2 : 72
Option 3 : 40
Option 4 : 30
Answer : A
Explanation : No. of ways of selecting atleast 1 red ball
= Selecting 3 R
+ Selecting 2 R and 1 B or G
+ 1 R and 2 B or G
Ways of selecting 3 Red balls
= 4C3
Ways of selecting 2R and 1 B or G
= 4C2*5C1
Ways of selecting 1R and 2 B or G
= 4C1*5C2
Required no. of ways
= 4C3 + (4C2*5C1) + (4C1*5C2)
= 4+30+40
= 74

Question25 : In how many different ways can the letters of the word LITTLE be arranged?
Option 1 : 360
Option 2 : 180
Option 3 : 720
Option 4 : 90
Answer : B
Explanation : The word LITTLE has 6 letters. (2L, 1I, 2T, 1E)

No. of ways
= 6!/(2!*1!*2!*1!)
= 180

Question26 : There are 10 yes or no questions. In how many ways can these be answered?
Option 1 : 1024
Option 2 : 256
Option 3 : 100
Option 4 : 20
Answer : A
Explanation : No. of ways of answering 1 question
= 2 (Y,N)

No. of ways of answering 2 question
= 2*2 (YY,YN,NY,NN)

No. of ways of answering 10 questions = 2^10 = 1024

Question27 : In how many ways can the letters of the word POCKET be arranged without changing the positions of the vowels?
Option 1 : 720
Option 2 : 120
Option 3 : 60
Option 4 : 24
Answer : D
Explanation : Without changing the positions of the two vowels, only the four consonants P,C,K,T have to be rearranged.

No. of ways of arranging 4 letters
= 4!
= 24

Question28 : In how many ways can the word APTITUDE be arranged such that the vowels always come together?
Option 1 : 8!
Option 2 : 1440
Option 3 : 576
Option 4 : 120
Answer : B
Explanation : The word APTITUDE has 8 letters. (4 vowels, 2 Ts, 1 P and 1 D)

The 4 vowels (A,E,I,U) should always come together. They can be considered as a single letter. So the word can be considered as a 5-letter word.

No. of ways of arranging the 4 vowels
= 4! = 24

No. of ways of arranging a 5-letter word
= 5!

Since there are two Ts, no. of ways
= 5!/2!
= 60

No. of ways of arranging 'APTITUDE' so that vowels dont come together
= 60*24
= 1440

Question29 : Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Option 1 : 210
Option 2 : 1050
Option 3 : 25200
Option 4 : 21400
Answer : C
Explanation : No. of ways of selecting 3 consonants(out of 7) and 2 vowels(out of 4)
= 7C3 * 4C2
= 35*6
= 210

There are 210 groups, each having 3 consonants and 2 vowels. Each group has 5 letters.

No. of ways of arranging those 5 letters
= 5!
= 120

Required no. of ways
= 210*120
= 25200

Question30 : From a group consisting of 8 boys and 5 girls, 6 persons are to be selected to form a team so that at least 3 boys are there in the team. In how many ways can this be done?
Option 1 : 768
Option 2 : 1568
Option 3 : 560
Option 4 : 1548
Answer : B
Explanation : The team can consist of
3 boys, 3 girls
4 boys, 2 girls
5 boys, 1 girl
6 boys

No. of ways
= (8C3*5C3)+ (8C4*5C2)+ (8C5*5C1)+ (8C6)
= (56*10)+(70*10)+(56*5)+28
= 1568





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