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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:

There are 4 books on fairy tales, 5 novels and 3 plays. In how many ways can you arrange these so that books on fairy tales are together, novels are together and plays are together and in the order, books on fairytales, novels and plays ?

A) 12400 B) 17820
C) 17280 D) 12460
 
Answer & Explanation Answer: C) 17280

Explanation:

There are 4 books on fairy tales and they have to be put together. They can be arranged in 4! ways.

 

Similarly, there are 5 novels.They can be arranged in 5! ways.

 

And there are 3 plays.They can be arranged in 3! ways.

 

So, by the counting principle all of them together can be arranged in 4!´5!´3! ways = 17280

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7 14611
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 11399
Q:

Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?

A) 2400 B) 6400
C) 3600 D) 7200
 
Answer & Explanation Answer: C) 3600

Explanation:

Let the beds be numbered 1 to 7.

 

Case 1 : Suppose Anju is allotted bed number 1. 

Then, Parvin cannot be allotted bed number 2. 

So Parvin can be allotted a bed in 5 ways. 

After alloting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.

So, in this case the beds can be allotted in 5´5!ways = 600 ways.

 

Case 2 : Anju is allotted bed number 7. 

Then, Parvin cannot be allotted bed number 6 

As in Case 1, the beds can be allotted in 600 ways.

 

Case 3 : Anju is allotted one of the beds numbered 2,3,4,5 or 6. 

Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed. For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.

Therefore, Parvin can be allotted a bed in 4 ways in all these cases.

After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.

Therefore, in each of these cases, the beds can be allotted in 4´ 5! = 480 ways. 

The beds can be allotted in (2x 600 + 5 x 480)ways = (1200 + 2400)ways = 3600 ways

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15 14941
Q:

If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this be done?

A) 720 B) 360
C) 240 D) 740
 
Answer & Explanation Answer: B) 360

Explanation:

We have to find number of permutations of 4 objects out of 6 objects.

 

This number is 6P4= 360

 

Therefore, cards can be sent in 360 ways.

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8 7860
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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8 10776
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

 

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2 9586
Q:

How many 3-digit numbers can be formed with the digits 1,4,7,8 and 9 if the digits are not repeated ?

A) 60 B) 26
C) 50 D) 64
 
Answer & Explanation Answer: A) 60

Explanation:

Three digit number will have unit’s, ten’s and hundred’s place.

 

Out of 5 given digits any one can take the unit’s place.

 

This can be done in 5 ways. ...              (i)

 

After filling the unit’s place, any of the four remaining digits can take the ten’s place.

 

This can be done in 4 ways. ...              (ii)

 

After filling in ten’s place, hundred’s place can be filled from any of the three remaining digits.

 

This can be done in 3 ways. ... (iii) 

 

So,by counting principle, the number of 3 digit numbers = 5x 4 x 3 = 60

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0 3288
Q:

Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.

A) 17756 B) 17576
C) 12657 D) 12666
 
Answer & Explanation Answer: B) 17576

Explanation:

The first letter from the right can be chosen in 26 ways because there are 26 alphabets.

 

Having chosen this, the second letter can be chosen in 26 ways

 

The first two letters can chosen in 26 x 26 = 676 ways

 

Having chosen the first two letters, the third letter can be chosen in 26 ways.

 

All the three letters can be chosen in 676 x 26 =17576 ways.

 

It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.

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