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:

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A) 135 B) 63
C) 125 D) 64
 
Answer & Explanation Answer: B) 63

Explanation:

Required number of ways = (7C5*3C2) = (7C2*3C1) = 63

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50 44937
Q:

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

A) 360 B) 700
C) 720 D) 120
 
Answer & Explanation Answer: C) 720

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

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49 48900
Q:

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A) 1260 B) 1400
C) 1250 D) 1600
 
Answer & Explanation Answer: A) 1260

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways. 

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

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48 45534
Q:

When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

A) 620 B) 671
C) 625 D) 567
 
Answer & Explanation Answer: B) 671

Explanation:

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

 

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

 

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

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48 29238
Q:

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

A) 4050 B) 3600
C) 1200 D) 5040
 
Answer & Explanation Answer: D) 5040

Explanation:

'LOGARITHMS' contains 10 different letters.

 

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.

 

10P4

 

= 5040.

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43 42363
Q:

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.

A) 525 B) 535
C) 545 D) 555
 
Answer & Explanation Answer: B) 535

Explanation:

The number of points of intersection of 37 lines is C237. But 13 straight lines out of the given 37 straight lines pass through the same point A.

 

Therefore instead of getting C213 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C211 points, we get only one point B.

 

 Hence the number of intersection points of the lines is C237-C213-C211 +2 = 535

 

 

 

 

 

 

 

  

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Filed Under: Permutations and Combinations
Exam Prep: CAT , Bank Exams , AIEEE
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43 22766
Q:

How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4?

A) 120 B) 360
C) 240 D) 424
 
Answer & Explanation Answer: B) 360

Explanation:

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times.

 

 Number of 7 digit numbers = 7!3!×2! = 420

 

But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'.

 

=6!3!×2! = 60

 

Hence the required number of 7 digits numbers = 420 - 60 = 360

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Exam Prep: AIEEE , Bank Exams , CAT
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43 29732
Q:

How many necklace of 12 beads each can be made from 18 beads of different colours?

A) 18! B) 18! x 19!
C) 18!(6 x 24) D) 18! x 30
 
Answer & Explanation Answer: C) 18!(6 x 24)

Explanation:

Here clock-wise and anti-clockwise arrangements are same.

 

Hence total number of circular–permutations: 18P122*12 = 18!6*24

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35 16492