Permutation and Combination problems|Permutation and Combination|Aptitude-part-1

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Permutation and Combination problems, for Interview Preparations

Permutation and Combination problems, for Bank Exams(IBPS,SBI PO,SBI Clerk)

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Permutation and Combination Aptitude problems,for Aptitude Test preparation

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Permutation and Combination problems,for MBA Exams preparation

Q1 . In how many ways can the words of Alternate be arranged?

1. 9!

2. 8!

3. 9!/2!

4. 9!/(2!x2!)

Correct Answer is

9!/(2!x2!)

SOLUTION

Alternate is a 9 letter word

so there are 9! Ways of arranging it

But a and e are repeated twice

so total ways 9!

(2!x2!)

Q2 . How many 4 digit numbers can be formed from 0,1,2,3,4,5,6?

1. 120

2. 180

3. 90

4. 240

Correct Answer is

180

SOLUTION

Total digits are 7

the first digit cannot be 0

so number of numbers formed

6x6x5 four digit numbers = 180

Q3 . In how many ways can a class of 10 students be seated around a circular table?

1. 8!

2. 7!

3. 9!

4. 120

Correct Answer is

9!

SOLUTION

A circular arrangement of n objects is (n-1)!

for 10 students

(10-1)=9!

Q4 . In how many ways can a king of hearts be picked?

1. 52

2. 52!

3. 52C2

4. 51

Correct Answer is

52

SOLUTION

We are going to select one unique card from 52

the answer is 52C1 =52

Q5 . how many 4 digits numbers multiplies of 5 or 10 can be formed from 0-9?

1. 900

2. 480

3. 952

4. 999

Correct Answer is

952

SOLUTION

For a number to be multipe of 5, it should end with 0 or 5

number ending with 0(last number is 0) 9x8x7

number ending with last number 5 is 8x8x7 because 0 cannot take first digit

9x8x7+8x8x7

504+448=952

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Permutation and Combination problems,Permutation and Combination, Aptitude solved questions

Target Audience for Permutation and Combination Aptitude problems

Permutation and Combination problems, for Interview Preparations

Permutation and Combination problems, for Bank Exams(IBPS,SBI PO,SBI Clerk)

Permutation and Combination problems,for Government Exams(SSC CGL,SSC CHSL)

Permutation and Combination Aptitude problems,for Aptitude Test preparation

Permutation and Combination problems,for Railway exams(RRB)

Permutation and Combination problems, for Postal Exams

IAS Exam preparation(UPSC)

Permutation and Combination problems,for MBA Exams preparation

Q1 . In how many ways can the words of Alternate be arranged?

1. 9!

2. 8!

3. 9!/2!

4. 9!/(2!x2!)

Correct Answer is

9!/(2!x2!)

SOLUTION

Alternate is a 9 letter word

so there are 9! Ways of arranging it

But a and e are repeated twice

so total ways 9!

(2!x2!)

Correct Answer is

9!/(2!x2!)

SOLUTION

Alternate is a 9 letter word

so there are 9! Ways of arranging it

But a and e are repeated twice

so total ways 9!

(2!x2!)

Q2 . How many 4 digit numbers can be formed from 0,1,2,3,4,5,6?

1. 120

2. 180

3. 90

4. 240

Correct Answer is

180

SOLUTION

Total digits are 7

the first digit cannot be 0

so number of numbers formed

6x6x5 four digit numbers = 180

Correct Answer is

180

SOLUTION

Total digits are 7

the first digit cannot be 0

so number of numbers formed

6x6x5 four digit numbers = 180

Q3 . In how many ways can a class of 10 students be seated around a circular table?

1. 8!

2. 7!

3. 9!

4. 120

Correct Answer is

9!

SOLUTION

A circular arrangement of n objects is (n-1)!

for 10 students

(10-1)=9!

Correct Answer is

9!

SOLUTION

A circular arrangement of n objects is (n-1)!

for 10 students

(10-1)=9!

Q4 . In how many ways can a king of hearts be picked?

1. 52

2. 52!

3. 52C2

4. 51

Correct Answer is

52

SOLUTION

We are going to select one unique card from 52

the answer is 52C1 =52

Correct Answer is

52

SOLUTION

We are going to select one unique card from 52

the answer is 52C1 =52

Correct Answer is

952

SOLUTION

For a number to be multipe of 5, it should end with 0 or 5

number ending with 0(last number is 0) 9x8x7

number ending with last number 5 is 8x8x7 because 0 cannot take first digit

9x8x7+8x8x7

504+448=952

For more Permutation and Combination aptitude problems

Check out our other content

Permutation and Combination problems,Permutation and Combination, Aptitude solved questions

Target Audience for Permutation and Combination Aptitude problems

Permutation and Combination problems, for Interview Preparations

Permutation and Combination problems, for Bank Exams(IBPS,SBI PO,SBI Clerk)

Permutation and Combination problems,for Government Exams(SSC CGL,SSC CHSL)

Permutation and Combination Aptitude problems,for Aptitude Test preparation

Permutation and Combination problems,for Railway exams(RRB)

Permutation and Combination problems, for Postal Exams

IAS Exam preparation(UPSC)

Permutation and Combination problems,for MBA Exams preparation

Q1 . In how many ways can the words of Alternate be arranged? 1. 9! 2. 8! 3. 9!/2! 4. 9!/(2!x2!) Check AnswerCorrect Answer is 9!/(2!x2!) SOLUTIONAlternate is a 9 letter wordso there are 9! Ways of arranging it But a and e are repeated twice so total ways 9!(2!x2!)

Correct Answer is 9!/(2!x2!)

SOLUTIONAlternate is a 9 letter wordso there are 9! Ways of arranging it But a and e are repeated twice so total ways 9!(2!x2!)

Q2 . How many 4 digit numbers can be formed from 0,1,2,3,4,5,6? 1. 120 2. 180 3. 90 4. 240 Check AnswerCorrect Answer is 180 SOLUTIONTotal digits are 7 the first digit cannot be 0so number of numbers formed6x6x5 four digit numbers = 180 (adsbygoogle = window.adsbygoogle || []).push({});

Correct Answer is 180

SOLUTIONTotal digits are 7 the first digit cannot be 0so number of numbers formed6x6x5 four digit numbers = 180

Q3 . In how many ways can a class of 10 students be seated around a circular table? 1. 8! 2. 7! 3. 9! 4. 120 Check AnswerCorrect Answer is 9! SOLUTIONA circular arrangement of n objects is (n-1)! for 10 students(10-1)=9!

Correct Answer is 9!

SOLUTIONA circular arrangement of n objects is (n-1)! for 10 students(10-1)=9!

Q4 . In how many ways can a king of hearts be picked? 1. 52 2. 52! 3. 52C2 4. 51 Check AnswerCorrect Answer is 52 SOLUTIONWe are going to select one unique card from 52 the answer is 52C1 =52 (adsbygoogle = window.adsbygoogle || []).push({});

Correct Answer is 52

SOLUTIONWe are going to select one unique card from 52 the answer is 52C1 =52

Correct Answer is 952

SOLUTIONFor a number to be multipe of 5, it should end with 0 or 5number ending with 0(last number is 0) 9x8x7 number ending with last number 5 is 8x8x7 because 0 cannot take first digit9x8x7+8x8x7 504+448=952

For more Permutation and Combination aptitude problems

Check out our other content

Q1 . In how many ways can the words of Alternate be arranged?

1. 9!

2. 8!

3. 9!/2!

4. 9!/(2!x2!)

Correct Answer is

9!/(2!x2!)

SOLUTION

Alternate is a 9 letter word

so there are 9! Ways of arranging it

But a and e are repeated twice

so total ways 9!

(2!x2!)

Correct Answer is

9!/(2!x2!)

SOLUTION

Alternate is a 9 letter word

so there are 9! Ways of arranging it

But a and e are repeated twice

so total ways 9!

(2!x2!)

Q2 . How many 4 digit numbers can be formed from 0,1,2,3,4,5,6?

1. 120

2. 180

3. 90

4. 240

Correct Answer is

180

SOLUTION

Total digits are 7

the first digit cannot be 0

so number of numbers formed

6x6x5 four digit numbers = 180

Correct Answer is

180

SOLUTION

Total digits are 7

the first digit cannot be 0

so number of numbers formed

6x6x5 four digit numbers = 180

Q3 . In how many ways can a class of 10 students be seated around a circular table?

1. 8!

2. 7!

3. 9!

4. 120

Correct Answer is

9!

SOLUTION

A circular arrangement of n objects is (n-1)!

for 10 students

(10-1)=9!

Correct Answer is

9!

SOLUTION

A circular arrangement of n objects is (n-1)!

for 10 students

(10-1)=9!

Q4 . In how many ways can a king of hearts be picked?

1. 52

2. 52!

3. 52C2

4. 51

Correct Answer is

52

SOLUTION

We are going to select one unique card from 52

the answer is 52C1 =52

Correct Answer is

52

SOLUTION

We are going to select one unique card from 52

the answer is 52C1 =52

Correct Answer is

952

SOLUTION

For a number to be multipe of 5, it should end with 0 or 5

number ending with 0(last number is 0) 9x8x7

number ending with last number 5 is 8x8x7 because 0 cannot take first digit

9x8x7+8x8x7

504+448=952