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:

In how many different ways can the letters of the word 'RITUAL' be arranged?

A) 720 B) 5040
C) 360 D) 180
 
Answer & Explanation Answer: A) 720

Explanation:

The number of letters in the given word RITUAL = 6

Then, 

Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!

=> 6 x 5 x 4 x 3 x 2 x 1 = 720

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11 1821
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.

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18 3970
Q:

A card is drawn from a pack of 52 cards. What is the probability that either card is black or a king? 

A) 15/52 B) 17/26
C) 13/17 D) 15/26
 
Answer & Explanation Answer: D) 15/26

Explanation:

Number of cards in a pack of cards = 52

Number of black cards = 26

Number of king cards = 4 (2 Red, 2 Black)

 

Required, the probability that if a card is drawn either card is black or a king = 

2652 + 452 = 3052 = 1526

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10 1671
Q:

A box contains 2 blue balls, 3 green balls and 4 yellow balls. In how many ways can 3 balls be drawn from the box, if at least one green ball is to be included in the draw?

A) 48 B) 24
C) 64 D) 32
 
Answer & Explanation Answer: C) 64

Explanation:

Total number of balls = 2 + 3 + 4 = 9

Total number of ways 3 balls can be drawn from 9 = 9C3

No green ball is drawn = 9 - 3 = 6 = 6C3

Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn

= 9C3 - 6C3

= 9x8x7/3x2  -  6x5x4/3x2

= 84 - 20

= 64 ways.

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6 3102
Q:

How many 4-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) 60 B) 48
C) 36 D) 20
 
Answer & Explanation Answer: A) 60

Explanation:

Here given the required digit number is 4 digit.

It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.

x x x 5

The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.

 

Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.

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2 2744
Q:

In how many ways the word 'SCOOTY' can be arranged such that 'S' and 'Y' are always at two ends?

A) 720 B) 360
C) 120 D) 24
 
Answer & Explanation Answer: D) 24

Explanation:

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements = 4!/2! x 2! = 4!

= 4 x 3 x 2

= 24 ways.

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6 3514
Q:

In how many ways word of 'GLACIOUS' can be arranged such that 'C' always comes at end?

A) 3360 B) 5040
C) 720 D) 1080
 
Answer & Explanation Answer: B) 5040

Explanation:

Given word is GLACIOUS has 8 letters.

=> C is fixed in one of the 8 places

Then, the remaining 7 letters can be arranged in 7! ways = 5040.

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2 2942
Q:

From a group of 7 boys and 6 girls, five persons are to be selected to form a team, so that at least 3 girls are there in the team. In how many ways can it be done?

A) 427 B) 531
C) 651 D) 714
 
Answer & Explanation Answer: B) 531

Explanation:

Given in the question that, there are 7 boys and 6 girls. 

Team members = 5

Now, required number of ways in which a team of 5 having atleast 3 girls in the team = 

6C3  x 7C2  + 6C4 x 7C1 + 6C5= 6x5x43x2x1 x 7x62x1 + 6x5x4x34x3x2x1 x 7 + 6x5x4x3x25x4x3x2x1= 420 + 105 + 6= 531.

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5 3279