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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:

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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8 2851
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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2 2782
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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3 2384
Q:

A group consists of 4 couples in which each of the 4 boys have one girl friend.In how many ways they can be arranged in a straight line such that boys and girls occupies alternate positions?

Answer

Answer : 1152


 


Total positions are 8.


In that boys can be arranged in 4 places and girls can be arranged in 4 places and hence this can be done in 2 ways.


i.e => 4! x 4! x 2 = 24 x 24 x 2=1152.

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10 2379
Q:

What is the sum of all 3 digits number that can be formed using digits 0,1,2,3,4,5 with no repitition ?

A) 28450 B) 26340
C) 32640 D) 36450
 
Answer & Explanation Answer: C) 32640

Explanation:

We know that zero can't be in hundreds place. But let's assume that our number could start with zero.

 

The formula to find sum of all numbers in a permutation is

 

111 x no of ways numbers can be formed for a number at given position x sum of all given digits

 

No of 1 s depends on number of digits

 

So,the answer us

 

111 x 20 x (0+1+2+3+4+5) = 33300

 

We got 20 as follows. If we have 0 in units place we can form a number in 4*5 ways. This is for all numbers. So we have substituted 20 in formula.

 

Now, this is not the final answer because we have included 0 in hundreds place. so we have to remove the sum of all numbers that starts with 0.

 

This is nothing but the sum of all 2 digits numbers formed by 1 2 3 4 5. Because 0 at first place makes it a 2 digit number.

 

So the sum for this is 11 x 4 x (1+2+3+4+5).
=660

 

Hope u understood why we use 4. Each number can be formed in 4x1 ways

 

So, the final answer is 33300-660 = 32640

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2 2365
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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3 2316
Q:

In how many different ways can the letters of the word 'HAPPYHOLI' be arranged?

A) 89,972 B) 90,720
C) 72,000 D) 81,000
 
Answer & Explanation Answer: B) 90,720

Explanation:

The given word HAPPYHOLI has 9 letters

These 9 letters can e arranged in 9! ways.

But here in the given word letters H & P are repeated twice each

Therefore, Number of ways these 9 letters can be arranged is 

9!2! x 2! = 9 x 8 x 7 x 6 x 5 x 4 x 32 = 90,720 ways.

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7 2305
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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