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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 how many ways, can zero or more letters be selected form the letters AAAAA?

A) 5 B) 6
C) 2 D) 8
 
Answer & Explanation Answer: B) 6

Explanation:

Selecting zero'A's= 1

Selecting one 'A's = 1

Selecting two 'A's = 1

Selecting three 'A's = 1

Selecting four 'A's = 1

Selecting five 'A's = 1

=> Required number ofways =6

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0 5884
Q:

A class has 8 football players. A 5-member team and a captain will be selected out of these 8 players. How many different selections can be made ? 

A) 210 B) 168
C) 1260 D) 10!/6!
 
Answer & Explanation Answer: B) 168

Explanation:

we can select the 5 member team out of the 8 in 8C5 ways = 56 ways.

The captain can be selected from amongst the remaining 3 players in 3 ways.

Therefore, total ways the selection of 5 players and a captain can be made = 56x3 = 168 ways.

 

(or)

 

Alternatively, A team of 6 members has to be selected from the 8 players. This can be done in 8C6 or 28 ways.

Now, the captain can be selected from these 6 players in 6 ways.

Therefore, total ways the selection can be made is 28x6 = 168 ways.

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8 5881
Q:

First,second and third prizes are to be awarded at an engineering fair in which 13 exhibits have been entered.In how many different ways can the prizes be awarded?

A) 1736 B) 1716
C) 1216 D) 1346
 
Answer & Explanation Answer: B) 1716

Explanation:

13P3= 13!/10! = 1716

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1 5849
Q:

A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?

A) 126 B) 240
C) 120 D) 260
 
Answer & Explanation Answer: A) 126

Explanation:

There are 8 students and the maximum capacity of the cars together is 9.

 

We may divide the 8 students as follows

 

Case I: 5 students in the first car and 3 in the second Or

 

Case II: 4 students in the first car and 4 in the second

 

Hence,     in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

 

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

 

Therefore, the total number of ways in which 8 students can travel is

 

8C3+8C4 = 56 + 70 = 126.

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0 5802
Q:

How many six digit odd numbers can be formed from the digits 0, 2, 3, 5, 6, 7, 8, and 9 (repetition not allowed)?

A) 8640 B) 720
C) 3620 D) 4512
 
Answer & Explanation Answer: A) 8640

Explanation:

Let the 6 digits of the required 6 digit number be abcdef

Then, the number to be odd number the last digit must be odd digit i.e 3, 5, 7 or 9

The first digit cannot be ‘0’ => possible digits = 3, 5, 7, 2, 6, 8

Remaining 4 places can be of 6 x 5 x 4 x 3 ways

This can be easily understood by

Therefore, required number of ways = 6 x 6 x 5 x 4 x 3 x 4 = 36 x 20 x 12 = 720 x 12

= 8640 ways.

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10 5769
Q:

A Cricket team of 23 people all shake hands with each other exactly once. How many hand shakes occur ?

A) 142 B) 175
C) 212 D) 253
 
Answer & Explanation Answer: D) 253

Explanation:

The first person shakes hands with 22 different people, the second person also shakes hands with 22 different people, but one of those handshakes was counted in the 22 for the first person, so the second person actually shakes hands with 21 new people. The third person, 20 people, and so on...
So,
22 + 21 + 20 + 19 + 18 + 17 + 16 + 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= n(n+1)/2 = 22 x 23 /2 = 11 x 23 = 253.

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4 5738
Q:

From a deck of 52 cards, a 7 card hand is dealt.How many distinct hands are there if the hand must contain 2 spades and 3 diamonds ?

A) 7250100 B) 7690030
C) 7250000 D) 3454290
 
Answer & Explanation Answer: A) 7250100

Explanation:

There are 13 spades,we must include 2: 13C2

 

There are 13 diamonds,we must include 3: 13C3

 

Since we can't have more than 2 spades and 3 diamonds,the remaining 2 cards must be pulled out from the 26 remaining clubs and hearts : 26C2

 

Therefore,13C2*13C3*26C2 = 7250100

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0 5734
Q:

There are 7 men and 10 women on a committee selection pool.A committee consisting of President,Vice-President, and Treasurer is to be formed.How many ways can exactly two men be on the committe ?

A) 1200 B) 1240
C) 1260 D) 1620
 
Answer & Explanation Answer: C) 1260

Explanation:

There are 7C2 ways of selecting two men, and 10C1 ways of selecting a woman.Since each position in the committee is different,arrange the three people in 3! ways.

 

So, 7C2*10C1*3! =1260

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1 5729