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:

On the occasion of New Year, each student of a class sends greeting cards to the others. If there are 21 students in the class, what is the total number of greeting cards exchanged by the students?

A) 380 B) 420
C) 441 D) 400
 
Answer & Explanation Answer: B) 420

Explanation:

Given total number of students in the class = 21

 

So each student will have 20 greeting cards to be send or receive (21 - 1(himself))

 

Therefore, the total number of greeting cards exchanged by the students = 20 x 21 = 420.

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Q:

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

A) 1260 B) 210
C) 210 x 6! D) 1512
 
Answer & Explanation Answer: A) 1260

Explanation:

A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.

Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210×6= 1260

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0 5122
Q:

A box contains 5 green, 4 yellow and 5 white pearls. Four pearls are drawn at random. What is the probability that they are not of the same colour ?

A)  11/91

B)  4/11

C)  1/11

D) 90/91

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

Explanation:

Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14


= C414 ways = 14×13×12×114×3×2×1= 1001


Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)

  C15+ C44+ C15 = 5+1+5 =11

 P(E) = n(E)n(S)=111001=191  

 

 Required probability = 1-191=9091

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8 5121
Q:

How many such pair of letters are there in the word ‘TROUBLED’ which have as many letters between them in the word as they have between them in the English alphabet?

A) 2 B) 3
C) 4 D) 5
 
Answer & Explanation Answer: A) 2

Explanation:

Hence there are two pairs

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13 5057
Q:

If two cards are taken one after another without replacing from a pack of 52 cards. What is the probability for the two cards be Ace ?

A) 51/1221 B) 42/221
C) 1/221 D) 52/1245
 
Answer & Explanation Answer: C) 1/221

Explanation:

Total Combination of getting a card from 52 cards = C152

Because there is no replacement, so number of cards after getting first card= 51

Now, Combination of getting an another card= C151


Total combination of getting 2 cards from 52 cards without replacement= (C152×C151


There are total 4 Ace in stack. Combination of getting 1 Ace is = C14

 

Because there is no replacement, So number of cards after getting first Ace = 3


Combination of getting an another Ace = C13

Total Combination of getting 2 Ace without replacement=C14×C13

Now,Probability of getting 2 cards which are Ace =C14×C13C152×C151 = 1/221.

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5 4982
Q:

There are 3 sections with 5 questions each. If four questions are selected from each section, the chance of getting different questions is ?

A) 1000 B) 625
C) 525 D) 125
 
Answer & Explanation Answer: D) 125

Explanation:

Methods for selecting 4 questions out of 5 in the first section = 5 x 4 x 3 x 2 x 1/4 x 3 x 2 x 1 = 5. Similarly for other 2 sections also i.e 5 and 5


So total methods = 5 x 5 x 5 = 125.

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4 4966
Q:

If (1 × 2 × 3 × 4 ........ × n) = n!, then 15! - 14! - 13! is equal to ___?

A) 14 × 13 × 13! B) 15 × 14 × 14!
C) 14 × 12 × 12! D) 15 × 13 × 13!
 
Answer & Explanation Answer: D) 15 × 13 × 13!

Explanation:

15! - 14! - 13!

= (15 × 14 × 13!) - (14 × 13!) - (13!)

= 13! (15 × 14 - 14 - 1)

= 13! (15 × 14 - 15)

= 13! x 15 (14 - 1)

= 15 × 13 × 13!

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9 4949
Q:

Find the number of ways to take 4 people and place them in groups of 3 at a time where order does not matter?

A) 4 B) 12
C) 36 D) 16
 
Answer & Explanation Answer: A) 4

Explanation:

Since order does not matter, use the combination formula 

C34 = 24/6 = 4

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