Permutations and Combinations Questions

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:

Find the number of ways to draw a straight, (suit does not matter) beginning with a 4 and ending with a 8?

A) 1024 B) 1296
C) 1094 D) 1200
 
Answer & Explanation Answer: A) 1024

Explanation:

There are 5 slots.

 

                   __ __ __ __ __

 

The first slot must be a four. There are 4 ways to put a four in the first slot.

 

There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.

 

(4)(4)(4)(4)(4) = 1024

 

Therefore there are 1024 different ways to produce the desired hand of cards.

Report Error

View Answer Report Error Discuss

0 4156
Q:

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

A) 11670 B) 12000
C) 11760 D) 20050
 
Answer & Explanation Answer: C) 11760

Explanation:

Required number of ways= 8C5×10C6 = (8C3×10C4)= 11760.

Report Error

View Answer Report Error Discuss

1 4070
Q:

36 identical books must be arranged in rows with the same number of books in each row. Each row must contain at least three books and there must be at least three rows. A row is parallel to the front of the room. How many different arrangements are possible ?

A) 5 B) 6
C) 7 D) 8
 
Answer & Explanation Answer: A) 5

Explanation:

The following arrangements satisfy all 3 conditions.

Arrangement 1: 3 books in a row; 12 rows.

Arrangement 2: 4 books in a row; 9 rows.
Arrangement 3: 6 books in a row; 6 rows.
Arrangement 4: 9 books in a row; 4 rows.
Arrangement 5: 12 books in a row; 3 rows.

 

Therefore, the possible arrangements are 5.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

8 4059
Q:

A committee of 4 people is to be formed from a group of 9 people.How many possible committees can be formed?

A) 120 B) 162
C) 126 D) 170
 
Answer & Explanation Answer: C) 126

Explanation:

This question is a combination since order is not important.

 

Answer = 7C3 = 126

Report Error

View Answer Report Error Discuss

1 4055
Q:

How many four digits numbers greater than 6000 can be made using the digits 0, 4, 2, 6 together with repetition.

A) 64 B) 63
C) 62 D) 60
 
Answer & Explanation Answer: B) 63

Explanation:

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

Report Error

View Answer Report Error Discuss

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

18 4040
Q:

On the occasion of New Year, each student of a class sends greeting cards to the others. If there are 21 students in the class, what is the total number of greeting cards exchanged by the students?

A) 380 B) 420
C) 441 D) 400
 
Answer & Explanation Answer: B) 420

Explanation:

Given total number of students in the class = 21

 

So each student will have 20 greeting cards to be send or receive (21 - 1(himself))

 

Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.

Report Error

View Answer Report Error Discuss

11 3974
Q:

If two cards are taken one after another without replacing from a pack of 52 cards. What is the probability for the two cards be Ace ?

A) 51/1221 B) 42/221
C) 1/221 D) 52/1245
 
Answer & Explanation Answer: C) 1/221

Explanation:

Total Combination of getting a card from 52 cards = C152

Because there is no replacement, so number of cards after getting first card= 51

Now, Combination of getting an another card= C151


Total combination of getting 2 cards from 52 cards without replacement= (C152×C151


There are total 4 Ace in stack. Combination of getting 1 Ace is = C14

 

Because there is no replacement, So number of cards after getting first Ace = 3


Combination of getting an another Ace = C13

Total Combination of getting 2 Ace without replacement=C14×C13

Now,Probability of getting 2 cards which are Ace =C14×C13C152×C151 = 1/221.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk , Analyst

5 3944
Q:

The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?

A) 2580 B) 3687
C) 4320 D) 5460
 
Answer & Explanation Answer: C) 4320

Explanation:

Let 'Y' be the youngest player.

The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways.

The required number of ways = 3(3!) = 4320.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations
Exam Prep: AIEEE , Bank Exams , CAT , GATE
Job Role: Bank Clerk , Bank PO

5 3920