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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:

12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?

A) 500 B) 490
C) 495 D) 540
 
Answer & Explanation Answer: C) 495

Explanation:

For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is 12C4 = 495.

 

Therefore, we can draw 495 quadrilaterals

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3 15643
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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8 15453
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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12 15444
Q:

There are three rooms in a Hotel: one single, one double and one for four persons. How many ways are there to house seven persons in these rooms ?

A) 105 B) 7! x 6!
C) 7!/5! D) 420
 
Answer & Explanation Answer: A) 105

Explanation:

Choose 1 person for the single room & from the remaining choose 2 for the double room & from the remaining choose 4 people for the four person room, 

 Then, 7C1 x 6C2 x 4C4 

= 7 x 15 x 1 = 105

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14 15362
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 15316
Q:

A group consists of 4 men, 6 women and 5 children. In how many ways can 2 men , 3 women and 1 child selected from the given group ?

A) 600 B) 610
C) 609 D) 599
 
Answer & Explanation Answer: A) 600

Explanation:

Two men, three women and one child can be selected in ⁴C₂ x ⁶C₃ x ⁵C₁ ways = 600 ways.

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6 15257
Q:

In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together?

A) 1440 B) 720
C) 2250 D) 3600
 
Answer & Explanation Answer: D) 3600

Explanation:

Given word is THERAPY.

Number of letters in the given word = 7

These 7 letters can be arranged in 7! ways.

Number of vowels in the given word = 2 (E, A)

The number of ways of arrangement in which vowels come together is 6! x 2! ways

 

Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways.

 

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12 15244
Q:

In how many ways can the letters of the word "PROBLEM" be rearranged to make 7 letter words such that none of the letters repeat?

A) 49 B) 7!
C) 7^7 D) 7^3
 
Answer & Explanation Answer: B) 7!

Explanation:

There are seven positions to be filled.

 

The first position can be filled using any of the 7 letters contained in PROBLEM.

 

The second position can be filled by the remaining 6 letters as the letters should not repeat.

 

The third position can be filled by the remaining 5 letters only and so on.

 

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

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