Sawaal

 Permutations and Combinations Questions

Popular Latest Rated Important Formulae

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:

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.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

7 11123
Q:

From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?

A) 91 B) 104
C) 109 D) 98
 
Answer & Explanation Answer: A) 91

Explanation:

We first count the number of committee in which

(i). Mr. Y is a member
(ii). the ones in which he is not

Case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member.
Now we are left with (6-1) men and (4-2) ladies (Mrs. X is not willing to join).

We can choose 1 more in5+2C1=7 ways.

Case (ii): If Mr. Y is not a member then we left with (6+4-1) people.
we can select 3 from 9 in 9C3=84 ways.

Thus, total number of ways is 7+84= 91 ways.

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

10 11025
Q:

In a G - 20 meeting there were total 20 people representing their own country. All the representative sat around a circular table. Find the number of ways in which we can arrange them around a circular table so that there is exactly one person between two representatives namely Manmohan and Musharraf.

A) 2 x (17!) B) 2 x (18!)
C) (3!) x (18!) D) (17!)
 
Answer & Explanation Answer: B) 2 x (18!)

Explanation:

A person can be chosen out of 18 people in 18 ways to be seated between Musharraf and Manmohan. Now consider Musharraf, Manmohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Musharraf and Manmohan can also be arranged in 2 ways. 

 

Required number of permutations = 18 x (17!) x 2 = 2 x 18!

Report Error

View Answer Report Error Discuss

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

2 10838
Q:

In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?

A) 4!/2! B) 3!/2!
C) (4! x 3!) / 2! D) 36
 
Answer & Explanation Answer: C) (4! x 3!) / 2!

Explanation:

ABACUS is a 6 letter word with 3 of the letters being vowels.

 

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

 

These 4 elements can be rearranged in 4! Ways.

 

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.

 

Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

3 10732
Q:

How many 3 letters words can be formed using the letters of the words hexagon?

A) 120 B) 210
C) 160 D) 200
 
Answer & Explanation Answer: B) 210

Explanation:

Since the word hexagon contains 7 different letters,the number of permutations is 7P3 = 7 x 6 x 5 =210

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

4 10682
Q:

A Group consists of 4 couples in which each of the  4 persons have one wife each. In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions?

A) 1152 B) 1278
C) 1296 D) None of these
 
Answer & Explanation Answer: A) 1152

Explanation:

Case I :  MW  MW  MW  MW

 

Case II:  WM  WM  WM  WM

 

Let us arrange 4 men in 4! ways, then we arrange 4 women in 4P4 ways at 4 places either left of the men or right of the men. Hence required number of arrangements

 

   4! × 4P4  + 4! × 4P4  = 2 × 576 = 1152

Report Error

View Answer Report Error Discuss

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

4 10439
Q:

Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel ?from A to E?

A) 36 B) 64
C) 74 D) 72
 
Answer & Explanation Answer: D) 72

Explanation:

 

The bus fromA to B can be selected in 3 ways.

The bus from B to C can be selected in 4 ways.

The bus from C toD can be selected in 2 ways.

The bus fromD to E can be selected in 3 ways.

So, by the General Counting Principle, one can travel fromA to E in 3 x 4 x 2 x 3 ways = 72

 

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

2 10169
Q:

How many ways are there to select a subcommittee of 7 members from among a committee of 17?

A) 19000 B) 19448
C) 19821 D) 19340
 
Answer & Explanation Answer: B) 19448

Explanation:

Since it does not matter what order the committee members are chosen in, the combination formula is used.

 

Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation.

 

C(17,7)= 19,448

Report Error

View Answer Report Error Discuss

Filed Under: Permutations and Combinations

1 9904