Sawaal

 Permutations and Combinations Questions

Popular Latest Rated Important Formulae

FACTS  AND  FORMULAE  FOR  PERMUTATIONS  AND  COMBINATIONS  QUESTIONS

 

 

1.  Factorial Notation: Let n be a positive integer. Then, factorial n, denoted n! is defined as: n!=n(n - 1)(n - 2) ... 3.2.1.

Examples : We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

 

2.  Permutations: The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex1 : All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

Ex2 : All permutations made with the letters a, b, c taking all at a time are:( abc, acb, bac, bca, cab, cba)

Number of Permutations: Number of all permutations of n things, taken r at a time, is given by:

Prn=nn-1n-2....n-r+1=n!n-r!

 

Ex : (i) P26=6×5=30   (ii) P37=7×6×5=210

Cor. number of all permutations of n things, taken all at a time = n!.

Important Result: If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,

such that p1+p2+...+pr=n

Then, number of permutations of these n objects is :

n!(p1!)×(p2! ).... (pr!)

 

3.  Combinations: Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Ex.1 : Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note that AB and BA represent the same selection.

Ex.2 : All the combinations formed by a, b, c taking ab, bc, ca.

Ex.3 : The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Ex.4 : Various groups of 2 out of four persons A, B, C, D are : AB, AC, AD, BC, BD, CD.

Ex.5 : Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combinations of n things, taken r at a time is:

Crn=n!(r !)(n-r)!=nn-1n-2....to r factorsr!

 

Note : (i)Cnn=1 and C0n =1     (ii)Crn=C(n-r)n

 

Examples : (i) C411=11×10×9×84×3×2×1=330      (ii)C1316=C(16-13)16=C316=560

Q:

From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels ?

A) 7600 B) 7200
C) 6400 D) 3600
 
Answer & Explanation Answer: B) 7200

Explanation:

From 5 consonants, 3 consonants can be selected in 5C3 ways.

 

From 4 vowels, 2 vowels can be selected in 4C2 ways.

 

Now with every selection, number of ways of arranging 5 letters is 5P5ways.

 

Total number of words = 5C3*4C2*5P5

 

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

16 27343
Q:

The number of ways in which 8 distinct toys can be distributed among 5 children?

A) 5P8 B) 5^8
C) 8P5 D) 8^5
 
Answer & Explanation Answer: B) 5^8

Explanation:

As the toys are distinct and not identical,

For each of the 8 toys, we have three choices as to which child will receive the toy. Therefore, there are 58 ways to distribute the toys.

 

Hence, it is 58 and not 85.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT , GATE , GRE , TOEFL
Job Role: Analyst , Bank Clerk , Bank PO , IT Trainer

16 15332
Q:

How many arrangements can be made out of the letters of the word DRAUGHT, the vowels never beings separated?

A) 1440 B) 720
C) 360 D) 640
 
Answer & Explanation Answer: A) 1440

Explanation:

There are 7 letters in the word DRAUGHT, the two vowels are A and U. Since, the vowels are not to be separated, AU can be considered as one entity. Therefore, the number of letters will be 6 instead of 7. The permutations will be P(6,6) = 6! ways.

 

But the two vowels A and U can be arranged in two ways, i.e. AU and UA. The required number of arrangements = 2!.6! = 1440 ways.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

15 14704
Q:

There are 7 non-collinear points. How many triangles can be drawn by joining these points?

A) 10 B) 30
C) 35 D) 60
 
Answer & Explanation Answer: C) 35

Explanation:

A triangle is formed by joining any three non-collinear points in pairs.

 

There are 7 non-collinear points

 

The number of triangles formed = 7C3 = 35

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

15 21350
Q:

There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?

A) 1170 B) 1200
C) 720 D) 360
 
Answer & Explanation Answer: A) 1170

Explanation:

Possibilities     Bowlers      Batsmen         Number of ways

 

                         6               9

 

         1              4                7              (6C4*9C7)

 

         2              5                6              6C5*9C6

 

         3              6                5              6C6*9C5

 

6C4*9C7 = 15 x 36 = 540

 

6C5*9C6 = 6 x 84 = 504

 

 6C6*9C5= 1 x 126 = 126

 

Total = 1170

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

15 17085
Q:

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

A) 2400 B) 6400
C) 3600 D) 7200
 
Answer & Explanation Answer: C) 3600

Explanation:

Let the beds be numbered 1 to 7.

 

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

 

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

 

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

15 14947
Q:

Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11} having 4 elements.

A) 340 B) 370
C) 320 D) 330
 
Answer & Explanation Answer: D) 330

Explanation:

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

 

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

 

This can be done in 11C4 ways = 330 ways

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

15 16138
Q:

How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters ?

A) 5^10 B) 10^5
C) 5P5 D) 5C5
 
Answer & Explanation Answer: A) 5^10

Explanation:

Each of the 10 letters can be posted in any of the 5 boxes.

 

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.

 

i.e. 5*5*5*….*5 (upto 10 times) = 5 ^ 10.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

14 22929