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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 there are 15 dots on a circle,how many triangles can be formed?

A) 455 B) 450
C) 469 D) 500
 
Answer & Explanation Answer: A) 455

Explanation:

There are 15 dots in total,and to make a triangle we need to select any three of those dots.

 

So,15C3 = 455

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5 12977
Q:

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together

A) 50000 B) 40500
C) 5040 D) 50400
 
Answer & Explanation Answer: D) 50400

Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

 

Thus, we have CRPRTN (OOAIO).

 

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

 

Number of ways arranging these letters =7!/2!= 2520.

 

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 3!/5!= 20 ways.

 

Required number of ways = (2520 x 20) = 50400.

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1 12454
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 12312
Q:

A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow ?

A) 3/17 B) 4/21
C) 2/21 D) 5/17
 
Answer & Explanation Answer: C) 2/21

Explanation:

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.

 

Probability that both marbles are blue = ³C₂/¹⁵C₂ = (3 x 2)/(15 x 14) = 1/35

Probability that both are yellow = ²C₂/¹⁵C₂ = (2 x 1)/(15 x 14) = 1/105

Probability that one blue and other is yellow = (³C₁ x ²C₁)/¹⁵C₂ = (2 x 3 x 2)/(15 x 14) = 2/35

 

Required probability = 1/35 + 1/105 + 2/35 = 3/35 + 1/105 = 1/35(3 + 1/3) = 10/(3 x 35) = 2/21

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13 12250
Q:

In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together?

A) 2800 B) 2880
C) 2600 D) 3980
 
Answer & Explanation Answer: B) 2880

Explanation:

They have to be arranged in the following way :

                                                                    L  T  L  T  L  T  L  T  L

The 5 lions should be arranged in the 5 places marked ‘L’.

This can be done in 5! ways.

The 4 tigers should be in the 4 places marked ‘T’.

This can be done in 4! ways.

Therefore, the lions and the tigers can be arranged in 5!´ 4! ways = 2880 ways.

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5 12216
Q:

A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

A) 123 B) 113
C) 246 D) 945
 
Answer & Explanation Answer: C) 246

Explanation:

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking. 

 

(i) 1 lady out of 4 and 4 gentlemen out of 6 

(ii) 2 ladies out of 4 and 3 gentlemen out of 6 

(iii) 3 ladies out of 4 and 2 gentlemen out of 6 

(iv) 4 ladies out of 4 and 1 gentlemen out of 6 

 

In case I the number of ways = C14×C46 = 4 x 15 = 60 

In case II the number of ways = C24×C36 = 6 x 20 = 120 

In case III the number of ways = C34×C26 = 4 x 15 = 60

In case IV the number of ways = C44×C16 = 1 x 6 = 6 

 

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

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10 11798
Q:

There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?

A) 18! x 19! B) 2 x 19!
C) 2 x 18! D) 18! x 18!
 
Answer & Explanation Answer: C) 2 x 18!

Explanation:

fix one person and the brothers B1 P B2 = 2 ways to do so.
other 17 people= 17!

 
Each person out of 18 can be fixed between the two=18, thus, 2 x 17! x 18=2 x 18!

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5 11557
Q:

Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?

A) 360 B) 780
C) 720 D) 240
 
Answer & Explanation Answer: C) 720

Explanation:

We have to arrange 6 books.

The number of permutations of n objects is n! = n. (n – 1) . (n – 2) ... 2.1

Here n = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720

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