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:

Find the value of 'n' for which the nth term of two AP'S:

15,12,9.... and -15,-13,-11...... are equal?

A) n = 2 B) n = 5
C) n = 29/5 D) n = 1
 
Answer & Explanation Answer: C) n = 29/5

Explanation:

Given are the two AP'S:

15,12,9.... in which a=15, d=-3.............(1) 

-15,-13,-11..... in which a'=-15 ,d'=2.....(2)

 

now using the nth term's formula,we get

a+(n-1)d = a'+(n-1)d'

substituting the value obtained in eq. 1 and 2,

15+(n-1) x (-3) = -15+(n-1) x 2

=> 15 - 3n + 3 = -15 + 2n - 2

=> 12 - 3n = -17 + 2n

=> 12+17 = 2n+3n

=> 29=5n

=> n= 29/5

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19 74614
Q:

In How many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?

A) 131 B) 191
C) 68 D) 3720
 
Answer & Explanation Answer: B) 191

Explanation:

Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind.

Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191

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19 12975
Q:

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

A) 28 B) 14
C) 21 D) 7
 
Answer & Explanation Answer: A) 28

Explanation:

Let x, y, z be the number of balls received by the three persons, then

  

x5, y5, z5 and x+y+z=21

 

Let u0, v0, w0 then

 

 x + y + z =21

 

 u + 5 + v + 5 + w + 5 = 21

 

 u + v + w = 6 

 

Total number of solutions = C3-16+3-1=C28=28

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18 8599
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 4708
Q:

How many arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are to occupy only odd places.

A) 720 B) 576
C) 567 D) 625
 
Answer & Explanation Answer: B) 576

Explanation:

There are 7 letters in the word Bengali of these 3 are vowels and 4 consonants.

 

There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in 4P3 ways and 4 constants can be arranged in 4P4 ways.

 

Number of words =4P3  x 4P4= 24 x 24 = 576

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18 22797
Q:

If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?

A) 32 B) 24
C) 72 D) 36
 
Answer & Explanation Answer: A) 32

Explanation:

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

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17 25635
Q:

A decision committee of 5 members is to be formed out of 4 Actors, 3 Directors and 2 Producers. In how many ways a committee of 2 Actors, 2 Directors and 1 Producer can be formed ?

A) 18 B) 24
C) 36 D) 32
 
Answer & Explanation Answer: C) 36

Explanation:

Required Number of ways = 4C2 × 3C2 × 2C1 = 36 = 36

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17 3766
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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