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 Permutations and Combinations Questions

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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:

There are 3 sections with 5 questions each. If four questions are selected from each section, the chance of getting different questions is ?

A) 1000 B) 625
C) 525 D) 125
 
Answer & Explanation Answer: D) 125

Explanation:

Methods for selecting 4 questions out of 5 in the first section = 5 x 4 x 3 x 2 x 1/4 x 3 x 2 x 1 = 5. Similarly for other 2 sections also i.e 5 and 5


So total methods = 5 x 5 x 5 = 125.

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4 4451
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 4441
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.

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0 4395
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.

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8 4352
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.

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5 4343
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

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1 4326
Q:

If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is 

A) 9! B) 9!/2
C) 10! D) 10!/2
 
Answer & Explanation Answer: D) 10!/2

Explanation:

As A1 speaks always after A2, they can speak only in  1st  to 9th places and 

 

A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place 

 

A2 can speak in 9 places the remaining 

 

 A3, A4, A5,...A10  has no restriction. So, they can speak in 9.8! ways. i.e

 

when A2 speaks in the first place, the number of ways they can speak is 9.8!.

 

When A2 speaks in second place, the number of ways they can speak is  8.8!.

 

When A2 speaks in third place, the number of ways they can speak is  7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8!

 

 

 

Therefore,Total Number of ways they can  speak = (9+8+7+6+5+4+3+2+1) 8! = 92(9+1)8! = 10!/2

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5 4323
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 4291