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1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A. 24400

B. 21300

C. 210

D. 25200

Here is the answer and explanation

Answer : Option D

Explanation :

Number of ways of selecting 3 consonants out of 7 = 7C3
Number of ways of selecting 2 vowels out of 4 = 4C2

Number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 = 7C3 x 4C2

$MF#%= \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}\right) \times \left(\dfrac{4 \times 3}{2 \times 1}\right) = 210 $MF#%

It means that we can have 210 groups where each group contains total 5 letters(3 consonants
and 2 vowels).

Number of ways of arranging 5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1 = 120

Hence, Required number of ways = 210 x 120 = 25200



2. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

A. 159

B. 209

C. 201

D. 212

Here is the answer and explanation

Answer : Option B

Explanation :

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.

Hence we have 4 choices as given below

We can select 4 boys ------(Option 1).
Number of ways to this = 6C4

We can select 3 boys and 1 girl ------(Option 2)
Number of ways to this = 6C3 x 4C1

We can select 2 boys and 2 girls ------(Option 3)
Number of ways to this = 6C2 x 4C2

We can select 1 boy and 3 girls ------(Option 4)
Number of ways to this = 6C1 x 4C3

Total number of ways
= (6C4) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C3)
= (6C2) + (6C3 x 4C1) + (6C2 x 4C2) + (6C1 x 4C1) [Applied the formula nCr = nC(n - r) ]

$MF#%= \left[\dfrac{6 \times 5 }{2 \times 1}\right] + \left[\left(\dfrac{6 \times 5 \times 4 }{3 \times 2 \times 1}\right) \times 4\right] + \left[\left(\dfrac{6 \times 5 }{2 \times 1}\right)\left(\dfrac{4 \times 3 }{2 \times 1}\right)\right] + \left[6 \times 4 \right] $MF#%

= 15 + 80 + 90 + 24
= 209



3. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

A. 624

B. 702

C. 756

D. 812

Here is the answer and explanation

Answer : Option C

Explanation :

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.

Hence we have the following 3 choices

We can select 5 men ------(Option 1)
Number of ways to do this = 7C5

We can select 4 men and 1 woman ------(Option 2)
Number of ways to do this = 7C4 x 6C1

We can select 3 men and 2 women ------(Option 3)
Number of ways to do this = 7C3 x 6C2

Total number of ways
= 7C5 + [7C4 x 6C1] + [7C3 x 6C2]
= 7C2 + [7C3 x 6C1] + [7C3 x 6C2] [Applied the formula nCr = nC(n - r) ]

$MF#%= \left[\dfrac{7 \times 6 }{2 \times 1}\right] + \left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times 6 \right] +\left[\left( \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \right) \times \left( \dfrac{6 \times 5}{2 \times 1} \right) \right]$MF#%

= 21 + 210 + 525 = 756



4. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

A. 610

B. 720

C. 825

D. 920

Here is the answer and explanation

Answer : Option B

Explanation :

The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels
should always come together. Hence these three vowels can be grouped and considered as a
single letter. That is, PTCL(OIA).

Hence we can assume total letters as 5. and all these letters are different.
Number of ways to arrange these letters = 5! = [5 x 4 x 3 x 2 x 1] = 120

All The 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves = 3! = [3 x 2 x 1] = 6

Hence, required number of ways = 120 x 6 = 720



5. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?

A. 47200

B. 48000

C. 42000

D. 50400

Here is the answer and explanation

Answer : Option D

Explanation :

The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and
these 5 vowels should always come together. Hence these 5 vowels can be grouped
and considered as a single letter. that is, CRPRTN(OOAIO).

Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times and
rest of the letters are different.

Number of ways to arrange these letters = $MF#%\dfrac{7!}{2!} = \dfrac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = $MF#%2520

In the 5 vowels (OOAIO), 'O' occurs 3 and rest of the vowels are different.

$MF#%\text{Number of ways to arrange these vowels among themselves = }\dfrac{5!}{3!} = \dfrac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 20$MF#%

Hence, required number of ways = 2520 x 20 = 50400



6. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A. 1

B. 126

C. 63

D. 64

Here is the answer and explanation

Answer : Option C

Explanation :

We need to select 5 men from 7 men and 2 women from 3 women

Number of ways to do this
= 7C5 x 3C2
= 7C2 x 3C1 [Applied the formula nCr = nC(n - r) ]

$MF#%= \left(\dfrac{7 \times 6 }{2 \times 1}\right) \times 3
$MF#%

= 21 x 3 = 63



7. In how many different ways can the letters of the word 'MATHEMATICS' be arranged such that the vowels must always come together?

A. 9800

B. 100020

C. 120960

D. 140020

Here is the answer and explanation

Answer : Option C

Explanation :

The word 'MATHEMATICS' has 11 letters. It has the vowels 'A','E','A','I' in it and
these 4 vowels must always come together. Hence these 4 vowels can be grouped and
considered as a single letter. That is, MTHMTCS(AEAI).

Hence we can assume total letters as 8. But in these 8 letters, 'M' occurs 2 times,
'T' occurs 2 times but rest of the letters are different.

Hence,number of ways to arrange these letters = $MF#%\dfrac{8!}{(2!)(2!)}=\dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)}= 10080$MF#%

In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels are different.

$MF#%\text{Number of ways to arrange these vowels among themselves = }\dfrac{4!}{2!} = \dfrac{4 \times 3 \times 2 \times 1}{2 \times 1} = 12$MF#%

Hence, required number of ways = 10080 x 12 = 120960



8. There are 8 men and 10 women and you need to form a committee of 5 men and 6 women. In how many ways can the committee be formed?

A. 10420

B. 11

C. 11760

D. None of these

Here is the answer and explanation

Answer : Option C

Explanation :

We need to select 5 men from 8 men and 6 women from 10 women

Number of ways to do this
= 8C5 x 10C6
= 8C3 x 10C4 [Applied the formula nCr = nC(n - r) ]

$MF#%=\left(\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1}\right) \left(\dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}\right)$MF#%

= 56 x 210
= 11760



9. How many 3-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

A. 720

B. 420

C. None of these

D. 5040

Here is the answer and explanation

Answer : Option A

Explanation :

The word 'LOGARITHMS' has 10 different letters.

Hence, the number of 3-letter words(with or without meaning) formed by using these letters
= 10P3
= 10 x 9 x 8
= 720



10. In how many different ways can the letters of the word 'LEADING' be arranged such that the vowels should always come together?

A. None of these

B. 720

C. 420

D. 122

Here is the answer and explanation

Answer : Option B

Explanation :

The word 'LEADING' has 7 letters. It has the vowels 'E','A','I' in it and
these 3 vowels should always come together. Hence these 3 vowels can be grouped
and considered as a single letter. that is, LDNG(EAI).

Hence we can assume total letters as 5 and all these letters are different.
Number of ways to arrange these letters = 5! = 5 x 4 x 3 x 2 x 1 = 120

In the 3 vowels (EAI), all the vowels are different.
Number of ways to arrange these vowels among themselves = 3! = 3 x 2 x 1= 6

Hence, required number of ways = 120 x 6= 720



11. A coin is tossed 3 times. Find out the number of possible outcomes.

A. None of these

B. 8

C. 2

D. 1

Here is the answer and explanation

Answer : Option B

Explanation :

When a coin is tossed once, there are two possible outcomes - Head(H) and Tale(T)

Hence, when a coin is tossed 3 times, the number of possible outcomes
= 2 x 2 x 2 = 8

(The possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH, TTT )



12. In how many different ways can the letters of the word 'DETAIL' be arranged such that the vowels must occupy only the odd positions?

A. None of these

B. 64

C. 120

D. 36

Here is the answer and explanation

Answer : Option D

Explanation :

The word 'DETAIL' has 6 letters which has 3 vowels (EAI) and 3 consonants(DTL)

The 3 vowels(EAI) must occupy only the odd positions.
Let's mark the positions as (1) (2) (3) (4) (5) (6).
Now, the 3 vowels should only occupy the 3 positions marked as (1),(3) and (5) in any order.
Hence, number of ways to arrange these vowels = 3P3
= 3! = 3 x 2 x 1 = 6

Now we have 3 consonants(DTL) which can be arranged in the remaining 3 positions in any order
Hence, number of ways to arrange these consonants = 3P3
= 3! = 3 x 2 x 1 = 6

Total number of ways
= number of ways to arrange the vowels x number of ways to arrange the consonants
= 6 x 6 = 36



13. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw?

A. 64

B. 128

C. 32

D. None of these

Here is the answer and explanation

Answer : Option A

Explanation :

From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.

Hence we have 3 choices as given below

We can select 3 black balls --------------------------(Option 1)
We can select 2 black balls and 1 non-black ball------(Option 2)
We can select 1 black ball and 2 non-black balls------(Option 3)

Number of ways to select 3 black balls = 3C3

Number of ways to select 2 black balls and 1 non-black ball = 3C2 x 6C1

Number of ways to select 1 black ball and 2 non-black balls = 3C1 x 6C2

Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2)
= 1 + (3C1 x 6C1) + (3C1 x 6C2) [Applied the formula nCr = nC(n - r) ]

$MF#%= 1 + \left[3 \times 6 \right] + \left[3 \times \left(\dfrac{6 \times 5}{2 \times 1}\right) \right]$MF#%

= 1 + 18 + 45
= 64



14. In how many different ways can the letters of the word 'JUDGE' be arranged such that the vowels always come together?

A. None of these

B. 48

C. 32

D. 64

Here is the answer and explanation

Answer : Option B

Explanation :

The word 'JUDGE' has 5 letters. It has 2 vowels (UE) in it and these 2 vowels should
always come together. Hence these 2 vowels can be grouped and considered as a single
letter. That is, JDG(UE).

Hence we can assume total letters as 4 and all these letters are different.
Number of ways to arrange these letters = 4!= 4 x 3 x 2 x 1 = 24

In the 2 vowels (UE), all the vowels are different.
Number of ways to arrange these vowels among themselves = 2! = 2 x 1 = 2

Total number of ways = 24 x 2 = 48



15. In how many ways can the letters of the word 'LEADER' be arranged?

A. None of these

B. 120

C. 360

D. 720

Here is the answer and explanation

Answer : Option C

Explanation :

The word 'LEADER' has 6 letters.

But in these 6 letters, 'E' occurs 2 times and rest of the letters are different.

$MF#%\text{Hence,number of ways to arrange these letters = }\dfrac{6!}{2!}=\dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}= 360$MF#%



16. How many words can be formed by using all letters of the word 'BIHAR'?

A. 720

B. 24

C. 120

D. 60

Here is the answer and explanation

Answer : Option C

Explanation :

The word 'BIHAR' has 5 letters and all these 5 letters are different.

Total words formed by using all these 5 letters = 5P5 = 5!
= 5 x 4 x 3 x 2 x 1 = 120



17. How many arrangements can be made out of the letters of the word 'ENGINEERING' ?

A. 924000

B. 277200

C. None of these

D. 182000

Here is the answer and explanation

Answer : Option B

Explanation :

The word 'ENGINEERING' has 11 letters.

But in these 11 letters, 'E' occurs 3 times,'N' occurs 3 times, 'G' occurs 2 times,
'I' occurs 2 times and rest of the letters are different.

$MF#%\text{Hence,number of ways to arrange these letters = }\dfrac{11!}{(3!)(3!)(2!)(2!)}\\
= \left[\dfrac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(3 \times 2 \times 1)(2 \times 1)(2 \times 1)}\right]= 277200 $MF#%



18. How many 3 digit 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?

A. 20

B. 16

C. 8

D. 24

Here is the answer and explanation

Answer : Option A

Explanation :

A number is divisible by 5 if the its last digit is a 0 or 5

We need to find out how many 3 digit numbers can be formed from the 6 digits (2,3,5,6,7,9)
which are divisible by 5.

Since the 3 digit number should be divisible by 5, we should take the digit 5
from the 6 digits(2,3,5,6,7,9) and fix it at the unit place.
There is only 1 way of doing this

            1  

Since the number 5 is placed at unit place, we have now five digits(2,3,6,7,9) remaining.
Any of these 5 digits can be placed at tens place

       5    1  

Since the digits 5 is placed at unit place and another one digits is placed at tens place,
we have now four digits remaining. Any of these 4 digits can be placed at hundreds place.

  4    5    1  

Required Number of three digit numbers = 4 x 5 x 1 = 20



19. How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI' using each letter exactly once?

A. 720

B. 24

C. None of these

D. 120

Here is the answer and explanation

Answer : Option D

Explanation :

The word 'DELHI' has 5 letters and all these letters are different.

Total words (with or without meaning) formed by using all these
5 letters using each letter exactly once
= Number of arrangements of 5 letters taken all at a time
= 5P5 = 5! = 5 x 4 x 3 x 2 x 1 = 120



20. What is the value of 100P2 ?

A. 9801

B. 12000

C. 5600

D. 9900

Here is the answer and explanation

Answer : Option D

Explanation :

100P2 = 100 x 99 = 9900



21. In how many different ways can the letters of the word 'RUMOUR' be arranged?

A. None of these

B. 128

C. 360

D. 180

Here is the answer and explanation

Answer : Option D

Explanation :

The word 'RUMOUR' has 6 letters.

But in these 6 letters, 'R' occurs 2 times, 'U' occurs 2 times and rest of the letters are different.

$MF#%\text{Hence, number of ways to arrange these letters = }\dfrac{6!}{(2!)(2!)}=\dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)}= 180$MF#%



22. There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?

A. 3200

B. None of these

C. 2400

D. 3600

Here is the answer and explanation

Answer : Option D

Explanation :

We have 6 periods and need to organize 5 subjects such that each subject is allowed
at least one period.

In 6 periods, 5 can be organized in 6P5 ways.

Remaining 1 period can be organized in 5P1 ways.

Total number of arrangements
= 6P5 x 5P1
= (6 x 5 x 4 x 3 x 2 ) x (5)
= 720 x 5
= 3600



23. How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?

A. 720

B. 360

C. 1420

D. 1680

Here is the answer and explanation

Answer : Option D

Explanation :

The first two places can only be filled by 3 and 5 respectively and there is only 1 way
of doing this

Given that no digit appears more than once. Hence we have 8 digits remaining(0,1,2,4,6,7,8,9)
So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways

Total number of ways = 8P4 = 8 x 7 x 6 x 5 = 1680



24. An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?

A. 100

B. 80

C. 110

D. 64

Here is the answer and explanation

Answer : Option B

Explanation :

He has has 10 patterns of chairs and 8 patterns of tables

Hence, A chair can be arranged in 10 ways and
A table can be arranged in 8 ways

Hence one chair and one table can be arranged in 10 x 8 ways = 80 ways



25. 25 buses are running between two places P and Q. In how many ways can a person go from P to Q and return by a different bus?

A. None of these

B. 600

C. 576

D. 625

Here is the answer and explanation

Answer : Option B

Explanation :

He can go in any bus out of the 25 buses.
Hence He can go in 25 ways.

Since he can not come back in the same bus that he used for travelling,
He can return in 24 ways.

Total number of ways = 25 x 24 = 600



26. A box contains 4 red, 3 white and 2 blue balls. Three balls are drawn at random. Find out the number of ways of selecting the balls of different colours?

A. 62

B. 48

C. 12

D. 24

Here is the answer and explanation

Answer : Option D

Explanation :

1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways

Total number of ways
= 4C1 x 3C1 x 2C1
=4 x 3 x 2
= 24



27. A question paper has two parts P and Q, each containing 10 questions. If a student needs to choose 8 from part P and 4 from part Q, in how many ways can he do that?

A. None of these

B. 6020

C. 1200

D. 9450

Here is the answer and explanation

Answer : Option D

Explanation :

Number of ways to choose 8 questions from part P = 10C8
Number of ways to choose 4 questions from part Q = 10C4

Total number of ways
= 10C8 x 10C4
= 10C2 x 10C4 [Applied the formula nCr = nC(n - r) ]

$MF#%= \left(\dfrac{10 \times 9 }{2 \times 1}\right)\left(\dfrac{10 \times 9 \times 8 \times 7 }{4 \times 3 \times 2 \times 1}\right) $MF#%

=45 x 210
= 9450



28. In how many different ways can 5 girls and 5 boys form a circle such that the boys and the girls alternate?

A. 2880

B. 1400

C. 1200

D. 3212

Here is the answer and explanation

Answer : Option A

Explanation :

In a circle, 5 boys can be arranged in 4! ways

Given that the boys and the girls alternate.
Hence there are 5 places for girls which can be arranged in 5! ways

Total number of ways = 4! x 5! = 24 x 120 = 2880



29. Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?

A. 120

B. 720

C. 125

D. 729

Here is the answer and explanation

Answer : Option D

Explanation :

The first ring can be worn in any of the 3 fingers
=> There are 3 ways of wearing the first ring

Similarly each of the remaining 5 rings also can be worn in 3 ways

Hence total number of ways

$MF#%= 3 \times 3 \times 3 \times 3 \times 3 \times 6 = 3^6 = 729$MF#%



30. In how many ways can 5 man draw water from 5 taps if no tap can be used more than once?

A. None of these

B. 720

C. 60

D. 120

Here is the answer and explanation

Answer : Option D

Explanation :

1st man can draw water from any of the 5 taps
2nd man can draw water from any of the remaining 4 taps
3rd man can draw water from any of the remaining 3 taps
4th man can draw water from any of the remaining 2 taps
5th man can draw water from remaining 1 tap

5 4 3 2 1

Hence total number of ways = 5 x 4 x 3 x 2 x 1 = 120



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Comments(29)


srinivas 17 Aug 2014 8:36 PM
nice problems
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mjmb b 30 Jul 2014 10:15 PM
thank you
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JAY GORE 05 May 2014 11:12 PM
216,163,120.......
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venky 28 Sep 2013 8:19 AM
thank uuu

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wakeup 27 Sep 2013 7:46 PM
hey guys,

In how many 6 digit numbers can be  formed with numbers from 0 to 6, without repetition such that the number is divisible by the digit at its unit places. 
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Bhupendra 29 Sep 2013 12:04 PM
Amandeep is right. But I have a more simple solution

Total possible numbers without repetition (without any divisibility conditions)= 6.6.5.4.3.2 = 4320

But the numbers with unit digit as 0 are not divisible by 0
Count of such numbers = 6.5.4.3.2.1 = 720

When the last two digits are 14 or 34 or 54, the number is not divisible by 4
Count of such numbers = 4.4.3.2.3.1 = 288

So , total count of 6 digit numbers which can be  formed with numbers from 0 to 6,
without repetition such that the number is divisible by the digit at its unit places
= 4320 - 720 - 288 = 3312
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Akshay 29 Sep 2013 11:22 AM
We have total 7 digits 0,1,2,3,4,5,6,

With 0 at the unit place
None of the numbers formed with 0 at unit place are divisible by 0. (Division by 0 is not defined)

With 1 at the unit place
All numbers formed with 1 at unit place are divisible by 1.
Keeping 1 at unit place ---(1 way)
Any of the 5 digits at the left most place (5 way)   (because 0 can not be used at the left most place)
Any of the remaining 5 digits at the next right place (5 way)
Any of the remaining 4 digits at the next right place (4 way)
Any of the remaining 3 digits at the next right place (3 way)
Any of the remaining 2 digits at the next right place (2 way)
Total such numbers  = 5.5.4.3.2 = 600

With 2 at the unit place
All numbers formed with 2 at unit place are divisible by 2.
Similar to the above case, total such numbers =  600

With 3 at the unit place
All numbers formed with 3 at unit place are divisible by 3 because sum of all digits, 21 is divisible by 3
Similar to the above case, total such numbers =  600

With 4 at the unit place
If a number is divisible by 4, last two digits must be divisible by 4

Keeping 4 at the unit place ---(1 way)
we can keep 2 or 6 at the tens place --- (2 ways)   (0 can also be kept,this case is considered seperately)
Any of the 5 digits at the left most place (5 way)   (because 0 can not be used at the left most place)
Any of the remaining 4 digits at the next right place(4 way)  
Any of the remaining 3 digits at the next right place(3 way)  
Any of the remaining 2 digits at the next right place(2 way)  
Total ways = 1.2.5.4.3.2 = 240

Keeping 4 at the unit place ---(1 way)
we can keep 0 at the tens place --- (1 way)  
Any of the remaining 5 digits at the left most place (5 way)   
Any of the remaining 4 digits at the next right place(4 way)  
Any of the remaining 3 digits at the next right place(3 way)  
Any of the remaining 2 digits at the next right place(2 way)
Total ways = 1.1.5.4.3.2 = 120

Total count of numbers with 4 at the unit place and is divisble by 4 =  240 + 120 = 360

With 5 at the unit place
All numbers formed with 5 at unit place are divisible by 5
Similar to the initial case, total such numbers =  600

With 6 at the unit place
All numbers formed with 6 at unit place are divisible by 6 as it is even and sum of digits, 21are divisible by 3
Similar to the initial case, total such numbers =  600

Required number of six digit numbers  = 0 + 600 + 600 + 600 + 360 + 600 + 600 = 3360
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Amandeep Singh 29 Sep 2013 11:32 AM
There is a small error in your answer when you calculated numbers with With 4 at the unit place

Corrected version is given below

Keeping 4 at the unit place ---(1 way)
we can keep 2 or 6 at the tens place --- (2 ways)   (0 can also be kept,this case is considered seperately)
Any of the 4 digits at the left most place (4 way)   (because 0 can not be used at the left most place)
Any of the remaining 4 digits at the next right place(4 way)  
Any of the remaining 3 digits at the next right place(3 way)  
Any of the remaining 2 digits at the next right place(2 way)  
Total ways = 1.2.4.4.3.2 = 192

Keeping 4 at the unit place ---(1 way)
we can keep 0 at the tens place --- (1 way)  
Any of the remaining 5 digits at the left most place (5 way)   
Any of the remaining 4 digits at the next right place(4 way)  
Any of the remaining 3 digits at the next right place(3 way)  
Any of the remaining 2 digits at the next right place(2 way)
Total ways = 1.1.5.4.3.2 = 120

Total count of numbers with 4 at the unit place and is divisble by 4 =  192 + 120 = 312

Hence, Required number of six digit numbers  = 0 + 600 + 600 + 600 + 312 + 600 + 600 = 3312
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Preethi 02 Sep 2013 2:37 AM
Guess the solution for 29th problem is wrong.

Ring 1 can go in any of the three fingers, so it has 3 choices. 

Ring 2 can go in any of the three fingers but it has four choices. There is a finger, say F3, which contains the ring R1. Now, on F3, R2 has two choices it can go above R1 or below R1. So, the total number of choices for R2 is 4.


Ring 3 can go in any of the three fingers but it now has 5 choices.

Ring 4 can go in any of the three fingers but it will now have 6 choices.

Ring 5 can go in any of the three fingers but it will now have 7 choices.

Ring 6 can go in any of the three fingers but it will now have 8 choices.



So, the total number of way of distribution of rings is = 3 x 4 x 5 x 6 x 7 x 8 = 8! / 2!


This is essentially how the formula (n + r -1)! / (r-1)! is derived for n rings to r fingers.


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mamta 12 Sep 2013 4:57 PM
how we solve permutation and combination questions in easy way
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