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 certain marathon had 50 people running for first prize, second, and third prize.How many ways are there to correctly guess the first, second, and third place winners?

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

Explanation:

There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.

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0 6276
Q:

If each of the 8 teams in a league must play each other three times,how many games will be played?

A) 72 B) 84
C) 36 D) 79
 
Answer & Explanation Answer: B) 84

Explanation:

Since,each member of the league must meet every other member of the league.If they only played each other once,there would be 8C2 games.Since,each pairing of teams will occur three times,the answer will be triple.

 

Therfore, 8C2*3=84

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1 6231
Q:

A school has scheduled three volleyball games, two soccer games, and four basketball games. You have a ticket allowing you to attend three of the games. In how many ways can you go to two basketball games and one of the other events?

A) 25 B) 30
C) 50 D) 75
 
Answer & Explanation Answer: B) 30

Explanation:

Since order does not matter it is a combination. 

 

The word AND means multiply. 

 

Given 4 basketball, 3 volleyball, 2 soccer. 

 

We want 2 basketball games and 1 other event. There are 5 choices left. 

C(n,r) 

C(How many do you have, How many do you want) 

C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1) 

C(4,2) x C(5,1) = (6)(5) = 30

 

Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.

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1 6221
Q:

When Dr. Ram arrives in his dispensary, he finds 12 patients waiting to see him. If he can see only one patients at a time,find the number of ways, he can schedule his patients if 3 leave in disgust before Dr. Ram gets around to seeing them.

A) 479001600 B) 79833600
C) 34879012 D) 67800983
 
Answer & Explanation Answer: B) 79833600

Explanation:

There are 12 - 3 = 9 patients.They can be seen in 12P9 = 79833600 ways

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3 6182
Q:

In how many different ways  can 3 students be associated with 4 chartered accountants,assuming that each chartered accountant can take at most one student?

A) 12 B) 36
C) 24 D) 16
 
Answer & Explanation Answer: C) 24

Explanation:

Number of permutations = 4P3 = 24

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1 5998
Q:

A pizza can have 3 toppings out of possible 7 toppings.How many different pizza's can be made?

A) 49 B) 35
C) 27 D) 25
 
Answer & Explanation Answer: B) 35

Explanation:

There are 7 toppings in total,and by selecting 3,we will make different types of pizza.

 

This question is a combination since having a different order of toppings will not make a different pizza.

 

So,7C3 =35

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0 5986
Q:

How many arrangements of the word TRIGONAL can be made if only two vowels and three consonants are used?

A) 6300 B) 3600
C) 6400 D) 7200
 
Answer & Explanation Answer: B) 3600

Explanation:

First we need to choose two vowels 3C2 and then three consonants 5C3. Now that we have 5 letters required to make the word,arrange them in 5! ways.

 

So,3C2*5C3*5! = 3600

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1 5924
Q:

There are 5 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next two have 6 choices each?

A) 1112 B) 2304
C) 1224 D) 2426
 
Answer & Explanation Answer: B) 2304

Explanation:

Number of questions = 5
Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6


Reuired total number of sequences

= 4 x 4 x 4 x 6 x 6

= 2304.

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