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:

A student council of 5 members is to be formed from a selection pool of 6 boys and 8 girls.How many councils can have Jason on the council?

A) 715 B) 725
C) 419 D) 341
 
Answer & Explanation Answer: A) 715

Explanation:

If Jason is on th ecouncil,this reduces the selction pool to only 13 people,out of which we still need to select 4.

 

So, 13C4 = 715

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1 4987
Q:

Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

A) 2589860 B) 2598970
C) 2598960 D) 2430803
 
Answer & Explanation Answer: C) 2598960

Explanation:

When a hand of cards is dealt, the order of the cards does not matter. If you are dealt two kings, it does not matter if the two kings came with the first two cards or the last two cards. Thus cards are combinations. There are 52 cards in a deck and we want to know how many different ways we can put them in groups of five at a time when order does not matter. The combination formula is used.

C(52,5) = 2,598,960

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1 4809
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 4697
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 4646
Q:

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A) 54 B) 64
C) 63 D) 36
 
Answer & Explanation Answer: C) 63

Explanation:

Required number of ways = 7C5 × 3C2=63

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0 4619
Q:

Eight first class and six second class petty officers are on the board of the 56 club. In how many ways can the members elect, from the board, a president, a vice-president, a secretary, and a treasurer if the president and secretary must be first class petty officers and the vice-president and treasurer must be second class petty officers?

A) 1500 B) 1860
C) 1680 D) 1640
 
Answer & Explanation Answer: C) 1680

Explanation:

Since two of the eight first class petty officers are to fill two different offices, we write 8P2=56

 

Then, two of the six second class petty officers are to fill two different offices; thus, we write 6P2 =30

 

The principle of choice holds in this case; therefore, the members have 56 x 30 = 1680 ways to select the required office holders

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0 4492
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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11 4468
Q:

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

A) 5040 B) 3650
C) 4150 D) 2520
 
Answer & Explanation Answer: D) 2520

Explanation:

Total number of letters in the word ABYSMAL are 7

 

Number of ways these 7 letters can be arranged are 7! ways

 

But the letter is repeated and this can be arranged in 2! ways

 

Total number of ways arranging ABYSMAL = 7!/2! = 5040/2 = 2520 ways.

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