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:

Out of seven consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

A) 25200 B) 25000
C) 25225 D) 24752
 
Answer & Explanation Answer: A) 25200

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) =( C37×C24 ) = 210.

Number of groups, each having 3 consonants and 2 vowels =210

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = 120. 

Required number of words = (210 x 120) = 25200.

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Q:

In how many ways word of 'GLACIOUS' can be arranged such that 'C' always comes at end?

A) 3360 B) 5040
C) 720 D) 1080
 
Answer & Explanation Answer: B) 5040

Explanation:

Given word is GLACIOUS has 8 letters.

=> C is fixed in one of the 8 places

Then, the remaining 7 letters can be arranged in 7! ways = 5040.

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Q:

If each of the vowels in the word 'MEAT' is kept unchanged and each of the consonants is replaced by the previous letter in the English alphabet, how many four-lettered meaningful words can be formed with the new letters, using each letter only once in each word?

A) 3 B) 4
C) 1 D) 2
 
Answer & Explanation Answer: A) 3

Explanation:
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Q:

16 persons are participated in a party. In how many differentways can they host the seat in a circular table, if the 2particular persons are to be seated on either side of the host?

A) 16! × 2 B) 14! × 2
C) 18! × 2 D) 14!
 
Answer & Explanation Answer: B) 14! × 2

Explanation:

(16 – 2)! × 2 = 14! × 2

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Q:

How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

A) 60 B) 48
C) 36 D) 20
 
Answer & Explanation Answer: A) 60

Explanation:

Here given the required digit number is 4 digit.

It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.

x x x 5

The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.

 

Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.

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Q:

In a bag, there are 8 red, 7 blue and 6 green flowers. One of the flower is picked up randomly. What is the probability that it is neither red nor green ?

A) 13

B)821

C)621

D)2021

A) Option A B) Option B
C) Option C D) Option D
 
Answer & Explanation Answer: A) Option A

Explanation:

Total number of flowers = (8+7+6) = 21.

 

Let E = event that the flower drawn is neither red nor green. 

= event taht the flower drawn is blue. 

 

--> n(E)= 7 

--> P(E)=  721=13 

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Q:

The ratio of the members of blue to red balls in abag is constant. When there were 44 red balls, the number of blue balls was 36. If the number of blue balls is 54, how many red balls will be in the bag?

A) 68 B) 32
C) 64 D) 66
 
Answer & Explanation Answer: D) 66

Explanation:
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Q:

In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?

A) 1,51,200 ways. B) 5,04,020 ways
C) 72,000 ways D) None of the above
 
Answer & Explanation Answer: A) 1,51,200 ways.

Explanation:

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in 6P4 ways 

Remaining 7 letters can be arranged in 7!/3! x 2! ways

 

Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.

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