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 'ABYSMAL' be arranged ?

A) 5040 B) 3650
C) 4150 D) 2520
 
Answer & Explanation Answer: D) 2520

Explanation:

Total number of letters in the word ABYSMAL are 7

 

Number of ways these 7 letters can be arranged are 7! ways

 

But the letter is repeated and this can be arranged in 2! ways

 

Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

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13 4605
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 4597
Q:

Eight first class and six second class petty officers are on the board of the 56 club. In how many ways can the members elect, from the board, a president, a vice-president, a secretary, and a treasurer if the president and secretary must be first class petty officers and the vice-president and treasurer must be second class petty officers?

A) 1500 B) 1860
C) 1680 D) 1640
 
Answer & Explanation Answer: C) 1680

Explanation:

Since two of the eight first class petty officers are to fill two different offices, we write 8P2=56

 

Then, two of the six second class petty officers are to fill two different offices; thus, we write 6P2 =30

 

The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders

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0 4592
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 4571
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 4550
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 4525
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 4509
Q:

If (1 × 2 × 3 × 4 ........ × n) = n!, then 15! - 14! - 13! is equal to ___?

A) 14 × 13 × 13! B) 15 × 14 × 14!
C) 14 × 12 × 12! D) 15 × 13 × 13!
 
Answer & Explanation Answer: D) 15 × 13 × 13!

Explanation:

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! x 15 (14 - 1)

= 15 × 13 × 13!

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