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 the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?

A) 4! x 4! B) 5! x 5!
C) 4! x 5! D) 3! x 4!
 
Answer & Explanation Answer: C) 4! x 5!

Explanation:

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions.

The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways.

Hence, the total number of ways = 4! × 5!

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10 16009
Q:

In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

A) 36 B) 25
C) 42 D) 120
 
Answer & Explanation Answer: A) 36

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

 

Let us mark these positions as under: 

                                                      (1) (2) (3) (4) (5) (6) 

Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.  

Number of ways of arranging the vowels = 3P3 = 3! = 6.

 

Also, the 3 consonants can be arranged at the remaining 3 positions. 

Number of ways of these arrangements = 3P3 = 3! = 6. 

Total number of ways = (6 x 6) = 36.

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5 15999
Q:

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

A) 53400 B) 17610
C) 11760 D) 45000
 
Answer & Explanation Answer: C) 11760

Explanation:

Required number of ways = 8C5*10C6 =  8C3*10C4 = 11760

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15 15908
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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9 15782
Q:

How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can take more than 10 letters ?

A) 5^10 B) 10^5
C) 5P5 D) 5C5
 
Answer & Explanation Answer: A) 5^10

Explanation:

Each of the 10 letters can be posted in any of the 5 boxes.

 

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.

 

i.e. 5*5*5*….*5 (upto 10 times) = 5 ^ 10.

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7 14993
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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17 14516
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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21 14258
Q:

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most  ?

A) 215 B) 268
C) 254 D) 216
 
Answer & Explanation Answer: A) 215

Explanation:

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt.

 

Maximum number of unsuccessful attempts = 216 - 1 = 215.

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