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:

The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and atleast 4 bowlers?

A) 1024 B) 1900
C) 2000 D) 1092
 
Answer & Explanation Answer: D) 1092

Explanation:

We are to choose 11 players including 1 wicket keeper and 4 bowlers  or, 1 wicket keeper and 5 bowlers.

 

Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in 2C1*5C4*9C6 = 840

 

Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in 2C1*5C5*9C5 =252

 

Total number of ways of selecting the team = 840 + 252 = 1092

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

35 33543
Q:

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

A) 196 B) 186
C) 190 D) 200
 
Answer & Explanation Answer: B) 186

Explanation:

When at least 2 women are included.

 

The committee may consist of 3 women, 2 men : It can be done in  4C3*6C2  ways

 

or, 4 women, 1 man : It can be done in  4C4*6C1ways

 

or, 2 women, 3 men : It can be done in 4C2*6C3 ways.

 

Total number of ways of forming the committees

 

= 4C2*6C3+4C3*6C2+4C4*6C1

 

= 6 x 20 + 4 x 15 + 1x 6

 

= 120 + 60 + 6 =186

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

87 62915
Q:

In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?

A) 180 B) 220
C) 240 D) 160
 
Answer & Explanation Answer: A) 180

Explanation:

Number of ways of choosing 2 black pens from 5 black pens in 5C2 ways.

 

Number of ways of choosing 2 white pens from 3 white pens in 3C2 ways.

 

Number of ways of choosing 2 red pens from 4 red pens in 4C2ways.

 

By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in 10 x 3 x 6 =180 ways.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

12 16645
Q:

12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?

A) 500 B) 490
C) 495 D) 540
 
Answer & Explanation Answer: C) 495

Explanation:

For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is 12C4 = 495.

 

Therefore, we can draw 495 quadrilaterals

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

3 14892
Q:

How many arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are to occupy only odd places.

A) 720 B) 576
C) 567 D) 625
 
Answer & Explanation Answer: B) 576

Explanation:

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

 

There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4P3 ways and 4 constants can be arranged in 4P4 ways.

 

Number of words =4P3  x 4P4= 24 x 24 = 576

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

18 21659
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 arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are never together.

A) 120 B) 640
C) 720 D) 540
 
Answer & Explanation Answer: C) 720

Explanation:

There are 7 letters in the word ‘Bengali of these 3 are vowels and 4 consonants.

 

Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.

 

Total number of words = 5! x 3!= 120 x 6 = 720

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

6 13170
Q:

In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?

A) 18000 B) 17280
C) 17829 D) 18270
 
Answer & Explanation Answer: B) 17280

Explanation:

Let 4 girls be one unit and now there are 6 units in all.

 

They can be arranged in 6! ways.

 

In each of these arrangements 4 girls can be arranged in 4! ways. 

 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

22 24592