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:

To fill 8 vacancies there are 15 candidates of which 5 are from ST. If 3 of the vacancies are reserved for ST candidates while the rest are open to all, Find the number of ways in which the selection can be done ?

A) 7920 B) 74841
C) 14874 D) 10213
 
Answer & Explanation Answer: A) 7920

Explanation:

ST candidates vacancies can be filled by C35 ways = 10 

Remaining vacancies are 5 that are to be filled by 12 

=> C512= (12x11x10x9x8)/(5x4x3x2x1) = 792 

Total number of filling the vacancies = 10 x 792 = 7920

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10 3493
Q:

In how many ways can letter of the word RAILINGS arrange so that R and S always come together?

A) 1260 B) 2520
C) 5040 D) 1080
 
Answer & Explanation Answer: C) 5040

Explanation:

The number of ways in which the letters of the word RAILINGS can be arranged such that R & S always come together is

 

Count R & S as only 1 space or letter so that RS or SR can be arranged => 7! x 2!

 

But in the word RAILINGS, I repeated for 2 times => 7! x 2!/2! = 7! ways = 5040 ways.

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7 3468
Q:

In how many ways can 4 sit at a round table for a group discussions?

A) 9 B) 3
C) 6 D) 12
 
Answer & Explanation Answer: C) 6

Explanation:

We get this using the formula of circular permutations i.e., (4 - 1)! = 3! =6

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0 3407
Q:

How many words can be formed with or without meaning by using three letters out of k, l, m, n, o without repetition of alphabets.

A) 60 B) 120
C) 240 D) 30
 
Answer & Explanation Answer: A) 60

Explanation:

Given letters are k, l, m, n, o = 5

number of letters to be in the words = 3

Total number of words that can be formed from these 5 letters taken 3 at a time without repetation of letters = 5P3 ways.

 5P3 = 5 x 4 x 3 = 60 words.

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9 3398
Q:

From a group of 7 boys and 6 girls, five persons are to be selected to form a team, so that at least 3 girls are there in the team. In how many ways can it be done?

A) 427 B) 531
C) 651 D) 714
 
Answer & Explanation Answer: B) 531

Explanation:

Given in the question that, there are 7 boys and 6 girls. 

Team members = 5

Now, required number of ways in which a team of 5 having atleast 3 girls in the team = 

6C3  x 7C2  + 6C4 x 7C1 + 6C5= 6x5x43x2x1 x 7x62x1 + 6x5x4x34x3x2x1 x 7 + 6x5x4x3x25x4x3x2x1= 420 + 105 + 6= 531.

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5 3358
Q:

Using numbers from 0 to 9 the number of 5 digit telephone numbers that can be formed is

A) 1,00,000 B) 59,049
C) 3439 D) 6561
 
Answer & Explanation Answer: C) 3439

Explanation:

The numbers 0,1,2,3,4,5,6,7,8,9 are 10 in number while preparing telephone numbers any number can be used any number of times.

 

This can be done in 105ways, but '0' is there

 

So, the numbers starting with '0' are to be excluded is 94 numbers.

 

 Total 5 digit telephone numbers = 105- 94 = 3439

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5 3343
Q:

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there ?

A) 205 B) 194
C) 209 D) 159
 
Answer & Explanation Answer: C) 209

Explanation:

We may have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys).


Required number of ways = (C16× C34) + C26×C24  + (C36× C14) + (C46)  

= (24+90+80+15) 

= 209.

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7 3323
Q:

In a plane 8 points are colliner  out of 12 points, then the number of triangles we get with those 12 points is 

A) 20 B) 160
C) 164 D) 220
 
Answer & Explanation Answer: C) 164

Explanation:

For a triangle, we need 3 non-collinear points. So with 12 points (when all the 12 are such that any three non-collinear is12C3. But among them 8 points are collinear.

 

If all these 8 points are different we get 8C3 triangles as they are collinear.

 

In 12C3 triangles, we do not get 8C3 triangles

 

Therefore, The number of triangles we get = 12C3-8C3 = 164

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3 3302