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Name: Permutations and Combinations : Aptitude Quiz
Subject: Aptitude
Topic: Permutations and Combinations : Aptitude Quiz
Questions: 12 Objective Type Questions
Time Allowed: 30 Minutes
Language: English
Important for: SSC ( CGL, CHSL, GD etc), State PCS, UPSC, Railway, IBPS Clerk, IBPS RRB, SBI Clerk, Police Exam, Army, BSF, CRPF exams, state level, समूह ग आदि 
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 Question 1 of 12
1. Question
1 pointsThe value of is:
CorrectIncorrectUnattempted  Question 2 of 12
2. Question
1 pointsHow many 4letter words with or without meaning, can be formed out of the letters of the word, A ‘LOGARITHMS’, if repetition of letters is not allowed?
Correct‘LOGARITHEM’ Contains 10 different letters.
Required number of words = Number of arrangements of 10 Letters, taking 4 at a time
= 5040Incorrect‘LOGARITHEM’ Contains 10 different letters.
Required number of words = Number of arrangements of 10 Letters, taking 4 at a time
= 5040Unattempted‘LOGARITHEM’ Contains 10 different letters.
Required number of words = Number of arrangements of 10 Letters, taking 4 at a time
= 5040  Question 3 of 12
3. Question
1 pointsIn how many ways can the letters of the word ‘APPLE’ be arranged?
CorrectThe word ‘APPLE’ contains 5 letters, 1A, 2P, 1L and 1E.
Required number of ways =IncorrectThe word ‘APPLE’ contains 5 letters, 1A, 2P, 1L and 1E.
Required number of ways =UnattemptedThe word ‘APPLE’ contains 5 letters, 1A, 2P, 1L and 1E.
Required number of ways =  Question 4 of 12
4. Question
1 pointsHow many words can be formed by using all the letters of the word, ‘ALLAHABAD’?
CorrectThe word ‘ALLAHABAD’ contains 9 letters, namely 4A, 2L, 1H, 1B and 1D.
Requisite number of words = 7560.IncorrectThe word ‘ALLAHABAD’ contains 9 letters, namely 4A, 2L, 1H, 1B and 1D.
Requisite number of words = 7560.UnattemptedThe word ‘ALLAHABAD’ contains 9 letters, namely 4A, 2L, 1H, 1B and 1D.
Requisite number of words = 7560.  Question 5 of 12
5. Question
1 pointsHow many words can be formed from the letters of the word ‘SIGNATURE’ so that the vowels always come together?
CorrectThe word ‘SIGNATURE’ contains 9 different letters.
When the vowels IAUE are taken together, they can be supposed to form an entity, treated as the letter.
Then, the letters to be arranged are SGNTR (IAAUE).
These 6 letters to be arranged in = 720 ways
The vowels in the group (IAUE) can be arranged among themselves in = 24 ways
Required number of words = (720×24) = 17280IncorrectThe word ‘SIGNATURE’ contains 9 different letters.
When the vowels IAUE are taken together, they can be supposed to form an entity, treated as the letter.
Then, the letters to be arranged are SGNTR (IAAUE).
These 6 letters to be arranged in = 720 ways
The vowels in the group (IAUE) can be arranged among themselves in = 24 ways
Required number of words = (720×24) = 17280UnattemptedThe word ‘SIGNATURE’ contains 9 different letters.
When the vowels IAUE are taken together, they can be supposed to form an entity, treated as the letter.
Then, the letters to be arranged are SGNTR (IAAUE).
These 6 letters to be arranged in = 720 ways
The vowels in the group (IAUE) can be arranged among themselves in = 24 ways
Required number of words = (720×24) = 17280  Question 6 of 12
6. Question
1 pointsIn how many different ways can the letters of the word ‘RUMOUR’ be arranged?
CorrectThe word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1O
Required number of ways = =180IncorrectThe word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1O
Required number of ways = =180UnattemptedThe word ‘RUMOUR’ contains 6 letters, namely 2R, 2U, 1M and 1O
Required number of ways = =180  Question 7 of 12
7. Question
1 pointsHow many arrangements can be made out the letters of the word ‘ENGINEERING’?
CorrectThe word ‘ENGINEERING’ Contains 11 letters, namely 3E, 3N, 2G, 2I or 1R.
required number of arrangements = =277200IncorrectThe word ‘ENGINEERING’ Contains 11 letters, namely 3E, 3N, 2G, 2I or 1R.
required number of arrangements = =277200UnattemptedThe word ‘ENGINEERING’ Contains 11 letters, namely 3E, 3N, 2G, 2I or 1R.
required number of arrangements = =277200  Question 8 of 12
8. Question
1 pointsIn how many different ways can the letters of the word ‘LEADING’ be arranged in such a way that the vowels always come together?
CorrectThe word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
The, we have to arrange the letters LDNG (EAI).
Now, 5 letters can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120×6)= 720IncorrectThe word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
The, we have to arrange the letters LDNG (EAI).
Now, 5 letters can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120×6)= 720UnattemptedThe word ‘LEADING’ has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
The, we have to arrange the letters LDNG (EAI).
Now, 5 letters can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120×6)= 720  Question 9 of 12
9. Question
1 pointsIn how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together.
CorrectThe word ‘OPTICAL’ Contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form on letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120×6)= 720IncorrectThe word ‘OPTICAL’ Contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form on letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120×6)= 720UnattemptedThe word ‘OPTICAL’ Contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form on letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120×6)= 720  Question 10 of 12
10. Question
1 pointsIn how many different ways can the letters of the word ‘AUCTION’ be arranged in such a way that the vowels always came together?
CorrectThe word ‘AUTION’ has 7 different letters.
When the vowels AUIO are always together, they can be supposed to form one letters.
Then, we have to arrange the letters CTN (AUIO).
now, 4 letters can be arranged in 4! = 24 ways.
The vowels (AUTO) can be arranged among themselves in 4! =24 ways.
Required number of ways = (24×24) = 576IncorrectThe word ‘AUTION’ has 7 different letters.
When the vowels AUIO are always together, they can be supposed to form one letters.
Then, we have to arrange the letters CTN (AUIO).
now, 4 letters can be arranged in 4! = 24 ways.
The vowels (AUTO) can be arranged among themselves in 4! =24 ways.
Required number of ways = (24×24) = 576UnattemptedThe word ‘AUTION’ has 7 different letters.
When the vowels AUIO are always together, they can be supposed to form one letters.
Then, we have to arrange the letters CTN (AUIO).
now, 4 letters can be arranged in 4! = 24 ways.
The vowels (AUTO) can be arranged among themselves in 4! =24 ways.
Required number of ways = (24×24) = 576  Question 11 of 12
11. Question
1 pointsHow many 3digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
CorrectSince each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
tens place can be filled by any of the remaining 5 numbers.
So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1×5×4)=20IncorrectSince each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
tens place can be filled by any of the remaining 5 numbers.
So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1×5×4)=20UnattemptedSince each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
tens place can be filled by any of the remaining 5 numbers.
So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1×5×4)=20  Question 12 of 12
12. Question
1 pointsOut of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
CorrectNumber of ways of selection (3 consonants out of 7) and (2 vowels out of 4)
= \left ( ^{7}C_{3}\times^{4}C_{2}\right )= \left ( \frac{7\times6\times5}{3\times2\times1}\times\frac{4\times3}{2\times1} \right )= 210
Number of groups, each having 3 consonants and 2 vowels = 210
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves
=
Required number of words = = 25200IncorrectNumber of ways of selection (3 consonants out of 7) and (2 vowels out of 4)
= \left ( ^{7}C_{3}\times^{4}C_{2}\right )= \left ( \frac{7\times6\times5}{3\times2\times1}\times\frac{4\times3}{2\times1} \right )= 210
Number of groups, each having 3 consonants and 2 vowels = 210
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves
=
Required number of words = = 25200UnattemptedNumber of ways of selection (3 consonants out of 7) and (2 vowels out of 4)
= \left ( ^{7}C_{3}\times^{4}C_{2}\right )= \left ( \frac{7\times6\times5}{3\times2\times1}\times\frac{4\times3}{2\times1} \right )= 210
Number of groups, each having 3 consonants and 2 vowels = 210
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves
=
Required number of words = = 25200