# CRE 1 - Basics | Modern Math - Permutation & Combination

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In a class, there are 4 boys & 3 girls. In how many ways can a teacher chose a class representative?

Answer: 7

**Explanation** :

Out of 7 students 1 class representative has to be selected.

∴ Number of ways of selecting 1 student out of 7 = ^{7}C_{1} = 7 ways.

Hence, 7.

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**CRE 1 - Basics | Modern Math - Permutation & Combination**

In a class, there are 4 boys & 3 girls. In how many ways can a teacher chose a boy representative & a girl representative for the class?

Answer: 12

**Explanation** :

Out of 4 boys 1 representative has to be selected.

∴ Number of ways of selecting 1 boy out of 4 = ^{4}C_{1} = 4 ways.

Out of 3 girls 1 representative has to be selected.

∴ Number of ways of selecting 1 girl out of 3 = ^{3}C_{1} = 3 ways.

∴ Total number of ways of selecting 1 boy representative and 1 girl representative for the class = 4 × 3 = 12 ways.

Hence, 12.

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**CRE 1 - Basics | Modern Math - Permutation & Combination**

An amusement park has 4 green and 3 blue entry doors whereas 2 orange and 5 yellow exit doors. In how many ways can a person enter the amusement park?

Answer: 7

**Explanation** :

There are 4 green and 3 blue entry doors i.e., total 7 entry doors.

Out of total 7 entry doors, the person can enter from any 1 of the doors.

∴ Number of ways of entering the park = number of ways of selecting 1 door out of 7 = ^{7}C_{1} = 7.

Hence, 7.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

An amusement park has 4 green and 3 blue entry doors whereas 2 orange and 5 yellow exit doors. In how many ways can a person exit the amusement park?

Answer: 7

**Explanation** :

There are 2 orange and 5 yellow exit doors i.e., total 7 exit doors.

Out of total 7 exit doors, the person can leave from any 1 of the doors.

∴ Number of ways of exiting the park = number of ways of selecting 1 door out of 7 = ^{7}C_{1} = 7.

Hence, 7.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

An amusement park has 4 green and 3 blue entry doors whereas 2 orange and 5 yellow exit doors. In how many ways can a person enter and exit the amusement park?

Answer: 49

**Explanation** :

Number of ways of entering the park = 7.

Number of ways of leaving the park = 7.

∴ Number of ways of entering and leaving the park = 7 × 7 = 49.

Hence, 49.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

There are 2 different trains, 3 different flights & 5 different bus services to travel from Chennai to Bangalore. There are 5 different trains, 6 different flights & 7 different bus services to travel from Bangalore to Bhopal. In how many ways can a person travel from Chennai to Bhopal via Bangalore?

Answer: 180

**Explanation** :

We need to first go from Chennai → Bangalore.

There are 2 different trains, 3 different flights & 5 different bus services to travel from Chennai to Bangalore.

Out of 10 ways we need to choose 1 i.e., ^{10}C_{1} = 10 ways.

Now, need to from Bangalore. → Bhopal

There are 5 different trains, 6 different flights & 7 different bus services to travel from Bangalore to Bhopal.

Out of 18 ways we need to choose 1 i.e., ^{18}C_{1} = 18 ways.

∴ Total number of ways to go from Chennai → Bangalore → Bhopal = 10 × 18 = 180 ways.

Hence, 180.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can 6 people be seated in a row?

Answer: 720

**Explanation** :

Let the seats be numbered as 1 2 3 4 5 6.

First person can be seated on any of the 6 chairs i.e., in ^{6}C_{1} = 6 ways.

Now, out of the remaining 5 seats, the 2nd person can be seated in ^{5}C_{1} = 5 ways.

Similarly,

3^{rd} person can be seated in 4 ways.

4^{th} person can be seated in 3 ways.

5^{th} person can be seated in 2 ways.

6^{th} person can be seated in 1 way.

∴ Total number of ways of seating these 6 people in a row = 6 × 5 × 4 × 3 × 2 × 1 = 720 ways.

**Note**: number of ways of arranging n objects in a row is n!.

Here, 6 people can be arranged in a row in 6! = 720 ways.

Hence, 720.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can 4 people be seated on 6 chairs in a row?

Answer: 360

**Explanation** :

Let the chairs be numbered as 1 2 3 4 5 6.

First person can be chaired on any of the 6 chairs i.e., in ^{6}C_{1} = 6 ways.

Now, out of the remaining 5 chairs, the 2nd person can be chaired in ^{5}C_{1} = 5 ways.

Similarly,

3^{rd} person can be chaired in 4 ways.

4^{th} person can be chaired in 3 ways.

∴ Total number of ways of chairing these 6 people in a row = 6 × 5 × 4 × 3 = 360 ways.

Hence, 360.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can 4 out of 6 people be seated on 4 chairs in a row?

Answer: 360

**Explanation** :

We first need to select which 4 people are to be seated.

We can select 4 people out of 6 in ^{6}C_{4} = 6!/(4!×2!) = 15 ways.

Now, 4 people can be seated on 4 chairs in 4! = 24 ways.

∴ Total ways in which 4 out of 6 people be seated on 4 chairs in a row = 15 × 24 = 360.

Hence, 360.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can 2 students be selected for a quiz out of 4 eligible candidates?

Answer: 6

**Explanation** :

Number of ways of selecting 2 students out of 4 = ^{4}C_{2} = 4!/(2!×2!) = 6 ways.

Hence, 6.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

How many 3-digit numbers are there with distinct digits, each digit being odd?

- A.
60

- B.
45

- C.
125

- D.
90

Answer: Option 60

**Explanation** :

Odd digits are 1, 3, 5, 7 and 9 i.e., 5 digits are available.

We have to form a 3-digit number.

Unit’s digit can be filled by either 1 or 3 or 5 or 7 or 9 i.e., in 5 ways.

Hundred’s digit can be filled by remaining 4 numbers in 4 ways.

Thousand’s digit can be filled by remaining 3 numbers in 3 ways.

∴ Total number of 3-digit numbers that can be formed with each digit being odd = 5 × 4 × 3 = 60.

Hence, 60.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

How many 4-digit numbers can be made using the digits 0 to 7 without repetition?

Answer: 1470

**Explanation** :

Here 0 cannot come thousand’s place, hence

Number of ways of filling thousand’s place = 7 ways (digits 1 to 7)

Number of ways of filling hundred’s place = 7 ways (now digit 0 is also available)

Number of ways of filling ten’s place = 6 ways

Number of ways of filling unit’s place = 5 ways

∴ Total 4-digit numbers = 7 × 7 × 6 × 5 = 1470.

Hence, 1470.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In printing all the numbers from 100 to 1000, how many times will the digit 5 be printed?

- A.
3980

- B.
4000

- C.
3700

- D.
280

Answer: Option D

**Explanation** :

1000 does not have a digit as 5 hence, we have to find how many times will the digit 5 be printed from 100 to 999 i.e., in all 3-digit numbers.

Number of 3-digit numbers whose unit’s digit is 5 = 9 × 10 = 90.

Number of 3-digit numbers whose ten’s digit is 5 = 9 × 10 = 90.

Number of 3-digit numbers whose hundred’s digit is 5 = 10 × 10 = 100.

Hence, digit 5 will be printed a total of 90 + 90 + 100 = 280 times.

Hence, option (d).

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can the first three prizes be distributed to the winners of a race with 12 participants? Assume that no two people finished in the same time.

- A.
3!

- B.
1320

- C.
12!

- D.
None of these

Answer: Option 1320

**Explanation** :

1^{st} prize can go to any of the 12 children i.e., in 12 ways.

2^{nd} prize can go to any of the remaining 11 children i.e., in 11 ways.

3^{rd} prize can go to any of the remaining 10 children i.e., in 10 ways.

∴ Total number of ways first 3 prizes can be given = 12 × 11 × 10 = 1320.

Hence, 1320.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can Javier invite at least 1 of his 10 friends to a party?

- A.
10

- B.
55

- C.
1024

- D.
1023

Answer: Option D

**Explanation** :

Javier can any number of friends from 1 to 10.

∴ Number of ways of inviting at least 1 friend = ^{10}C_{1} + ^{10}C_{2} + ^{10}C_{3} + … + ^{10}C_{10} = 210 – 1 = 1023.

**Alternately,**

For each friend there are 2 possibilities i.e., either he will be invited or not.

Hence, total number of ways of inviting friends = 2^{10}.

This includes one possibility where no one is invited.

∴ Number of ways of inviting at least 1 friend = 2^{10} – 1 = 1023.

Hence, option (d).

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In a party of 10 people, every person met each other, shook their hands and exchanged visiting cards. Find the number of cards exchanged & handshakes in the party respectively.

- A.
45, 90

- B.
45, 45

- C.
90, 45

- D.
90, 90

Answer: Option A

**Explanation** :

2 people will shake hands once, hence number of handshakes is same as selecting a pair of people = ^{10}C_{2} = 45.

When 2 people shake hands both of them will give their card to the other person. Hence, for every pair of persons shaking hands their will be 2 cards exchanged.

∴ Number of cards exchanged = 45 × 2 = 90.

Hence, option (a).

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can a student answer each of the 5 multiple choice questions, each containing 4 answer options, at random? Assume that answering a question entails marking one of the 4 answer choices.

- A.
4

^{5} - B.
5

^{5} - C.
5

^{4} - D.
4

^{4}

Answer: Option A

**Explanation** :

We have to answer all the 5 questions.

Number of ways of answer a question is 4. (each question has 4 answer options)

∴ Number of ways of answering 5 questions is 4^{5}.

Hence, option (a).

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

In how many ways can a student answer at least 1 out of 6 multiple choice questions, each containing 5 answer options, at random? Assume that answering a question entail marking one of the 5 answer choices.

- A.
5

^{6}- 1 - B.
6

^{6}- 1 - C.
6

^{6} - D.
None of these

Answer: Option B

**Explanation** :

We have to answer at least 1 question.

Number of ways of answer / not answering a question is 6. (each question has 5 answer options + 1 way is not answering the question)

∴ Total Number of ways of answering / not answering 6 questions is 6^{6}.

Now out of all these 6^{6} possibilities, there would be 1 possibility where no answer is marked. Hence, we will have to subtract this possibility.

∴ Number of ways of answering at least 1 question = 6^{6} – 1.

Hence, option (b).

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

Find the number of ways in which 5 boys & 5 girls can be seated in a row such that no two boys are seated together.

Answer: 216000

**Explanation** :

Since no two boys should be together, we first make all 5 girls sit in a row in 5! = 120 ways.

| G | G | G | G | G |

Now, we have 6 places where boys can be seated.

We first select 5 places out of these 6 = ^{6}C_{2} = 15 ways.

now, arrange 5 boys in these 5 places = 5! = 120 ways,

∴ Total number of ways of seating boys in between girls = 15 × 120 = 1800 ways.

∴ Number of ways in which 5 boys & 5 girls can be seated in a row such that no two boys are seated together = 120 × 1800 = 216000.

Hence, 216000.

Workspace:

**CRE 1 - Basics | Modern Math - Permutation & Combination**

Find the number of ways in which 5 boys & 5 girls can be seated in a row such that all the boys are never together.

- A.
10! – 5!

- B.
10! - 6! × 5!

- C.
12! – 6!

- D.
None of these

Answer: Option B

**Explanation** :

Total number of ways of arranging these 10 people = 10!

Now let’s find out the number of ways when all boys are together.

Let’s call the group of 5 boys as X.

There will be 5! ways X can be grouped.

For each of these groups of boys, we now have 5 girls and X, i.e., 6 people to be arranged. This can be done in 6! ways.

∴ Number of ways when all boys are together = 6! × 5!.

∴ Number of ways in which 5 boys & 5 girls can be seated in a row such that all the boys are never together = 10! - 6! × 5!

Hence, option (b).

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