Problems on Permutations and Combinations - Solved Examples(Set 1)
1. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A. 24400B. 21300
C. 210D. 25200
| Discuss
answer with explanation

Answer: Option D

Explanation:

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

Number of ways of selecting 3 consonants from 7 and 2 vowels from 4
= 7C3 × 4C2
@@=\left(\dfrac{7 × 6 × 5}{3 × 2 × 1}\right) × \left(\dfrac{4 × 3}{2 × 1}\right) \\= 210 @@

It means 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!=5×4×3×2×1=120@@

Hence, required number of ways
@@=210×120=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. 159B. 209
C. 201D. 212
| Discuss
answer with 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 options 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 × 4C1

We can select 2 boys and 2 girls ...(option 3)
Number of ways to this = 6C2 × 4C2

We can select 1 boy and 3 girls ...(option 4)
Number of ways to this = 6C1 × 4C3

Total number of ways
= 6C4 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C3
= 6C2 + 6C3 × 4C1 + 6C2 × 4C2 + 6C1 × 4C1[∵ nCr = nC(n-r)]

@@=\dfrac{6×5}{2×1}+\dfrac{6×5×4}{3×2×1}×4@@ @@+\dfrac{6×5}{2×1}×\dfrac{4×3}{2×1}+6×4@@
@@=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 in the committee. In how many ways can it be done?
A. 624B. 702
C. 756D. 812
| Discuss
answer with 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 options.

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 × 6C1

We can select 3 men and 2 women ...(option 3)
Number of ways to do this = 7C3 × 6C2

Total number of ways
= 7C5 + (7C4 × 6C1) + (7C3 × 6C2)
= 7C2 + (7C3 × 6C1) + (7C3 × 6C2)[∵ nCr = nC(n - r) ]

@@= \dfrac{7×6}{2×1}+\dfrac{7×6×5}{3×2×1}×6@@ @@+\dfrac{7×6×5}{3×2×1}×\dfrac{6×5}{2×1}@@
@@=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. 610B. 720
C. 825D. 920
| Discuss
answer with 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!=5×4×3×2×1=120@@

All the 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves
@@=3!=3×2×1=6@@

Hence, required number of ways
@@=120×6=720@@


5. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
A. 47200B. 48000
C. 42000D. 50400
| Discuss
answer with 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
@@=\dfrac{7!}{2!}=\dfrac{7×6×5×4×3×2×1}{2×1}=2520@@

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

Number of ways to arrange these vowels among themselves @@=\dfrac{5!}{3!}=\dfrac{5×4×3×2×1}{3×2×1}=20@@

Hence, required number of ways
@@=2520×20=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. 1B. 126
C. 63D. 64
| Discuss
answer with 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 × 3C2
= 7C2 × 3C1 [∵ nCr = nC(n-r)]
@@=\dfrac{7×6}{2×1}×3\\=21×3=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. 9800B. 100020
C. 120960D. 140020
| Discuss
answer with 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
@@=\dfrac{8!}{(2!)(2!)}@@ @@=\dfrac{8×7×6×5×4×3×2×1}{(2×1)(2×1)}=10080@@

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

Number of ways to arrange these vowels among themselves @@=\dfrac{4!}{2!}=\dfrac{4×3×2×1}{2×1}=12@@

Hence, required number of ways
@@=10080×12=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. 10420B. 11
C. 11760D. None of these
| Discuss
answer with 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 × 10C6
= 8C3 × 10C4 [∵ nCr = nC(n-r)]
@@=\left(\dfrac{8×7×6}{3×2×1}\right)\left(\dfrac{10×9×8×7}{4×3×2×1}\right)\\=56×210
=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. 720B. 420
C. None of theseD. 5040
| Discuss
answer with 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
@@=10×9×8\\=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 theseB. 720
C. 420D. 122
| Discuss
answer with 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!=5×4×3×2×1=120@@

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

Hence, required number of ways
@@=120×6=720@@


11. A coin is tossed 3 times. Find out the number of possible outcomes.
A. None of theseB. 8
C. 2D. 1
| Discuss
answer with 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
@@=2×2×2=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 theseB. 64
C. 120D. 36
| Discuss
answer with 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!=3×2×1=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!=3×2×1=6@@

Total number of ways
= number of ways to arrange the vowels × number of ways to arrange the consonants
@@=6×6=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. 64B. 128
C. 32D. None of these
| Discuss
answer with 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 × 6C1
Number of ways to select 1 black ball and 2 non-black balls
= 3C1 × 6C2

Total number of ways
= 3C3 + 3C2 × 6C1 + 3C1 × 6C2
= 3C3 + 3C1 × 6C1 + 3C1 × 6C2[∵ nCr = nC(n-r)]
@@=1+3×6+3×\dfrac{6×5}{2×1}\\=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 theseB. 48
C. 32D. 64
| Discuss
answer with explanation

Answer: Option B

Explanation:

The word 'JUDGE' has 5 letters. It has 2 vowels (UE) 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!=4×3×2×1=24@@

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

Total number of ways @@=24×2=48@@


15. In how many ways can the letters of the word 'LEADER' be arranged?
A. None of theseB. 120
C. 360D. 720
| Discuss
answer with 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.

Hence,number of ways to arrange these letters
@@=\dfrac{6!}{2!}=\dfrac{6×5×4×3×2×1}{2×1}=360@@


16. How many words can be formed by using all letters of the word 'BIHAR'?
A. 720B. 24
C. 120D. 60
| Discuss
answer with explanation

Answer: Option C

Explanation:

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

Total number of words that can be formed by using all these 5 letters
= 5P5 @@=5!@@
@@=5×4×3×2×1=120@@


17. How many arrangements can be made out of the letters of the word 'ENGINEERING' ?
A. 924000B. 277200
C. None of theseD. 182000
| Discuss
answer with 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.

Hence,number of ways to arrange these letters

@@=\dfrac{11!}{(3!)(3!)(2!)(2!)}\\=\dfrac{11×10×9×8×7×6×5×4×3×2}{(3×2)(3×2)(2)(2)}\\=277200@@

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. 20B. 16
C. 8D. 24
| Discuss
answer with explanation

Answer: Option A

Explanation:

A number is divisible by 5 if the its last digit is 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

51

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

451

Required Number of three digit numbers
@@=4×5×1=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. 720B. 24
C. None of theseD. 120
| Discuss
answer with explanation

Answer: Option D

Explanation:

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

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


20. What is the value of 100P2 ?
A. 9801B. 12000
C. 5600D. 9900
| Discuss
answer with explanation

Answer: Option D

Explanation:

100P2 @@=100×99=9900@@


21. In how many different ways can the letters of the word 'RUMOUR' be arranged?
A. None of theseB. 128
C. 360D. 180
| Discuss
answer with explanation

Answer: Option D

Explanation:

The word 'RUMOUR' has 6 letters.

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

Hence, number of ways to arrange these letters
@@=\dfrac{6!}{(2!)(2!)}=\dfrac{6×5×4×3×2}{2×2}=180@@


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. 3200B. None of these
C. 1800D. 3600
| Discuss
answer with explanation

Answer: Option C

Explanation:

Solution 1

5 subjects can be arranged in 6 periods in 6P5 ways.

Any of the 5 subjects can be organized in the remaining period (5C1 ways).

Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.

Total number of arrangements
@@=\dfrac{~^{6}\text{P}_5 × ~^5\text{C}_1}{2!}=1800@@


Solution 2

5 subjects can be selected in 5C5 ways.

1 subject can be selected in 5C1 ways.

These 6 subjects can be arranged themselves in 6! ways.

Since two subjects are same, we need to divide by 2!

Therefore, total number of arrangements
@@=\dfrac{~^5\text{C}_5 × ~^5\text{C}_1 × 6!}{2!}=1800@@


Solution 3

Select any 5 periods (6C5 ways).
Allocate a different subject to each of these 5 periods (1 way).
These 5 subjects can be arranged themselves in 5! ways.
Select the 6th period (1 way).
Allocate a subject to this period (5C1 ways).
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Therefore, required number of ways
@@=\dfrac{~^6\text{C}_5 ×1×5!×1×~^5\text{C}_1}{2!}=1800@@


Solution 4

There are 5 subjects and 6 periods. Each subject must be allowed in at least one period. Therefore, two periods will have same subject and remaining four periods will have different subjects.

Select the two periods where the same subject is taught. This can be done in 6C2 ways.

Allocate a subject two these two periods(5C1 ways).

Remaining 4 subjects can be arranged in the remaining 4 periods in 4! ways.

Required number of ways
= 6C2 × 5C1 × 4! = 1800


23. How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?
A. 720B. 360
C. 1420D. 1680
| Discuss
answer with explanation

Answer: Option D

Explanation:

The first two places can only be filled by 3 and 5 respectively and there is only 1 way for 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 @@=8×7×6×5=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. 100B. 80
C. 110D. 64
| Discuss
answer with explanation

Answer: Option B

Explanation:

He has 10 patterns of chairs and 8 patterns of tables

A chair can be selected in 10 ways.
A table can be selected in 8 ways.

Hence one chair and one table can be selected in @@10×8@@ 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 theseB. 600
C. 576D. 625
| Discuss
answer with explanation

Answer: Option B

Explanation:

He can go in any of the 25 buses (25 ways).

Since he cannot come back in the same bus, he can return in 24 ways.

Total number of ways @@=25×24=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. 62B. 48
C. 12D. 24
| Discuss
answer with 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 × 3C1 × 2C1
@@=4×3×2\\=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 theseB. 6020
C. 1200D. 9450
| Discuss
answer with 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 × 10C4
= 10C2 × 10C4[∵ nCr = nC(n-r)]
@@=\left(\dfrac{10×9}{2×1}\right)\left(\dfrac{10×9×8×7}{4×3×2×1}\right)\\=45×210\\=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. 2880B. 1400
C. 1200D. 3212
| Discuss
answer with explanation

Answer: Option A

Explanation:

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

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

Total number of ways
@@=4!×5!=24×120=2880@@


29. Find out the number of ways in which 6 rings of different types can be worn in 3 fingers?
A. 120B. 720
C. 125D. 729
| Discuss
answer with explanation

Answer: Option D

Explanation:

The first ring can be worn in any of the 3 fingers (3 ways).

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

Hence total number of ways
@@=3×3×3×3×3×3=3^6=729@@


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 theseB. 720
C. 60D. 120
| Discuss
answer with 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.

54321

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

Set 1Set 2Set 3Set 4Set 5
 
 
 
 
Comments(68) Sign in (optional)
showing 1-10 of 68 comments,   sorted newest to the oldest
vijay
2015-06-24 01:56:35 
in how many ways sum s can be formed using exactly n variables......

input
5 -->(s)
2---->(n)

output
4

reason :
2+3 3+2 1+4  4+1

can anyone help me to find the logic?
(0) (0) Reply
Jay
2015-06-24 09:05:59 
You are looking for integer partitions. Refer Counting Integral Solutions
(0) (0) Reply
Kushal Vyas
2015-06-15 08:20:13 
why o!=1???
(0) (0) Reply
shuvam gupta
2015-06-23 06:44:35 
There are three proof:
A)Pattern
4!=5!/5
3!=4!/4
2!=3!/3
1!=2!/2
And
0!=1!/1===1...
B)Practical sums:
No.of ways of arranging
The word"APPLE" 
ANS
as there are 2 p and 1 a,l,e
So,5!/(2!*1!*1!*1!)
But other24 alphabets are not there so they are 0 in no.
For them it will be 0!
If 0!=1*0=0
So our answer must be undefined,but in reality it is not
So ,0!=1
C)Practical thinking:
If there are 3 items we can arrange in 3! Ways
If there are 2 items we can arrange in 2! Ways
If there are 1 items we can arrange in1! Ways
If there are 0 items we can arrange in 1 way...it present state...so0!=1.
(0) (0) Reply
Manu
2015-07-30 04:56:07 
Hai,Bro what u said is right and simply we can learn by C programming for 0! is easy way of understanding. Don't think bad its just an example for knowing the factorial case. Those who know C language it is easily understandable..
Ex:

#include<stdio.h>
main()
{
  int fact=1,i,num;
  printf("Enter the Number:\n");
  scanf("%d",&num);
  for(i=1;i<=num;i++)
  {
   fact=fact*i;
  }
  printf("%d\n",fact);
}

It can be understood simply that. The factorial of 0!=1 is always 1. If not the factorial case is wrong.
(0) (0) Reply
Andrea
2015-04-04 05:42:18 
How many words with or without dictionary meaning can be formed using the letters of the word EQUATION so the vowel and consonant are side by side?
(0) (0) Reply
alia
2015-09-16 11:51:51 
there are 5 vowels so 5! ways of arranging them.
3 consonants, so 3! ways of arranging them.
overall there are 2 ways to arrange the groups of vowels and consonants.

Hence the answer would be 2! x 5!x 3! = 1440
(0) (0) Reply
ajay gupta
2016-01-07 11:47:18 
no ..it will be the same as vowels comes together.. means 3 consonants and one group of vowels as asumed 4 and such that 4! ways and 5 vowels are arranged in 5! ways and hence = 4!*5!= 2880 ways
(0) (0) Reply
sam
2016-01-07 11:48:21 
"it will be the same as vowels comes together."

No. With your logic, QEUAIOTN is also a valid arrangement which is not. Vowels and consonants must be side by side, like 'EUAIO'QTN

Alia's answer is correct.
(0) (0) Reply
Dev
2015-04-06 16:22:57 
Assuming that all letters needs to be used.

EQUATION - 5 vowels and 3 consonants

5 vowels - can be arranged in 5! ways
3 consonants - can be arranged in 3! ways
each of this group can be arranged in 2! ways

Total number of ways = 5! * 3! * 2! = 1440
Required number of words = 1440
(0) (0) Reply
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