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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:

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

A) 601 B) 600
C) 603 D) 602
 
Answer & Explanation Answer: A) 601

Explanation:

If the word started with the letter A then the remaining 5 positions can be filled in  5! Ways.

 

If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

 

If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary.

 

The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

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83 43789
Q:

Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct

A) 65000 B) 64000
C) 72000 D) 36000
 
Answer & Explanation Answer: A) 65000

Explanation:

Out of 26 alphabets two distinct letters can be chosen in 26P2 ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. 

 

Combined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.

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3 8407
Q:

How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language?

A) 1000 B) 100
C) 500 D) 999
 
Answer & Explanation Answer: B) 100

Explanation:

1 million distinct 3 digit initials are needed.

 

Let the number of required alphabets in the language be ‘n’.

 

Therefore, using ‘n’ alphabets we can form n * n * n = n3 distinct 3 digit initials.

 

Note distinct initials is different from initials where the digits are different.

 

For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.

 

This n3 different initials = 1 million 

i.e. n3=106  (1 million = 106)

  => n = 102 = 100

 

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

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7 10609
Q:

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A) 135 B) 63
C) 125 D) 64
 
Answer & Explanation Answer: B) 63

Explanation:

Required number of ways = (7C5*3C2) = (7C2*3C1) = 63

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50 45083
Q:

When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

A) 620 B) 671
C) 625 D) 567
 
Answer & Explanation Answer: B) 671

Explanation:

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

 

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

 

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

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48 29393
Q:

From 5 consonants and 4 vowels, how many words can be formed using 3 consonants and 2 vowels ?

A) 7600 B) 7200
C) 6400 D) 3600
 
Answer & Explanation Answer: B) 7200

Explanation:

From 5 consonants, 3 consonants can be selected in 5C3 ways.

 

From 4 vowels, 2 vowels can be selected in 4C2 ways.

 

Now with every selection, number of ways of arranging 5 letters is 5P5ways.

 

Total number of words = 5C3*4C2*5P5

 

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

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16 27336
Q:

There are 5 novels and 4 biographies. In how many ways can 4 novels and 2 biographies can be arranged on a shelf ?

A) 26100 B) 21600
C) 24000 D) 36000
 
Answer & Explanation Answer: B) 21600

Explanation:

4 novels can be selected out of 5 in 5C4 ways.

2 biographies can be selected out of 4 in 4C2 ways.

Number of ways of arranging novels and biographies = 5C4*4C2  = 30

After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.

By the Counting Principle, the total number of arrangements = 30 x 720 = 21600

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7 13733
Q:

If nC10=nC12  then,find n.

A) 10 B) 12
C) 22 D) 24
 
Answer & Explanation Answer: C) 22

Explanation:

Using, Crn=Cn-rn we get 

n – 10 = 12

or, n = 12 + 10 = 22

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19 13141