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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:

How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

A) 59 B) 56
C) 64 D) 55
 
Answer & Explanation Answer: A) 59

Explanation:

The first letter is E and the last one is R.

 

Therefore, one has to find two more letters from the remaining 11 letters.

 

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

 

The second and third positions can either have two different letters or have both the letters to be the same.

 

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

 

Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways.

 

Total number of possibilities = 56 + 3 = 59

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22 20911
Q:

Four ladies A, B, C and D and four gentlemen E, F, G and H are sitting in a circle round a table facing each other.

Directions:

(1) No two ladies or two gentlemen are sitting side by side.

(2) C, who is sitting between G and E is facing D.

(3) F is between D and A and is facing G.

(4) H is to the right of B.

Question:

1. Who are immediate neighbours of B?

2. E is facing whom?

A) G & H , H B) F & H , B
C) E & F , F D) E & H , G
 
Answer & Explanation Answer: A) G & H , H

Explanation:

From the directions given : 

f11482726547.png image

From the fig. it is clear that 

1) neighbours of B are G , H.

2) E is facing H.

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22 18053
Q:

In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together ?

A) 18000 B) 17280
C) 17829 D) 18270
 
Answer & Explanation Answer: B) 17280

Explanation:

Let 4 girls be one unit and now there are 6 units in all.

 

They can be arranged in 6! ways.

 

In each of these arrangements 4 girls can be arranged in 4! ways. 

 

Total number of arrangements in which girls are always together = 6! x 4!= 720 x 24 = 17280

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22 24544
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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21 27582
Q:

There are 5 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next two have 6 choices each?

A) 1112 B) 2304
C) 1224 D) 2426
 
Answer & Explanation Answer: B) 2304

Explanation:

Number of questions = 5
Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6


Reuired total number of sequences

= 4 x 4 x 4 x 6 x 6

= 2304.

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20 5165
Q:

Find the total numbers greater than 4000 that can be formed with digits 2, 3, 4, 5, 6 no digit being repeated in any number ?

A) 120 B) 256
C) 192 D) 244
 
Answer & Explanation Answer: C) 192

Explanation:

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated.

 

The number can be 4 digited but greater than 4000 or 5 digited.

 

Number of 4 digited numbers greater than 4000 are
4 or 5 or 6 can occupy thousand place => 3 x P34 = 3 x 24 = 72. 

 

5 digited numbers = P55 = 5! = 120 

So the total numbers greater than 4000 = 72 + 120 = 192

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20 13883
Q:

If nC10=nC12  then,find n.

A) 10 B) 12
C) 22 D) 24
 
Answer & Explanation Answer: C) 22

Explanation:

Using, Crn=Cn-rn we get 

n – 10 = 12

or, n = 12 + 10 = 22

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19 13101
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 10804