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 = 7C1 = 7 ways.
Hence, 7.
Workspace:
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 = 4C1 = 4 ways.
Out of 3 girls 1 representative has to be selected.
∴ Number of ways of selecting 1 girl out of 3 = 3C1 = 3 ways.
∴ Total number of ways of selecting 1 boy representative and 1 girl representative for the class = 4 × 3 = 12 ways.
Hence, 12.
Workspace:
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 = 7C1 = 7.
Hence, 7.
Workspace:
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 = 7C1 = 7.
Hence, 7.
Workspace:
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:
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., 10C1 = 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., 18C1 = 18 ways.
∴ Total number of ways to go from Chennai → Bangalore → Bhopal = 10 × 18 = 180 ways.
Hence, 180.
Workspace:
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 6C1 = 6 ways.
Now, out of the remaining 5 seats, the 2nd person can be seated in 5C1 = 5 ways.
Similarly,
3rd person can be seated in 4 ways.
4th person can be seated in 3 ways.
5th person can be seated in 2 ways.
6th 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:
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 6C1 = 6 ways.
Now, out of the remaining 5 chairs, the 2nd person can be chaired in 5C1 = 5 ways.
Similarly,
3rd person can be chaired in 4 ways.
4th 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:
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 6C4 = 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:
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 = 4C2 = 4!/(2!×2!) = 6 ways.
Hence, 6.
Workspace:
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:
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:
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:
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 :
1st prize can go to any of the 12 children i.e., in 12 ways.
2nd prize can go to any of the remaining 11 children i.e., in 11 ways.
3rd 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:
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 = 10C1 + 10C2 + 10C3 + … + 10C10 = 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 = 210.
This includes one possibility where no one is invited.
∴ Number of ways of inviting at least 1 friend = 210 – 1 = 1023.
Hence, option (d).
Workspace:
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 C
Explanation :
2 people will shake hands once, hence number of handshakes is same as selecting a pair of people = 10C2 = 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 (c).
Workspace:
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)
45
- (b)
55
- (c)
54
- (d)
44
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 45.
Hence, option (a).
Workspace:
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)
56 - 1
- (b)
66 - 1
- (c)
66
- (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 66.
Now out of all these 66 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 = 66 – 1.
Hence, option (b).
Workspace:
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: 86400
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 = 6C5 = 6 ways.
now, arrange 5 boys in these 5 places = 5! = 120 ways,
∴ Total number of ways of seating boys in between girls = 6 × 120 = 720 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 × 720 = 86400.
Hence, 86400.
Workspace:
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).
Workspace:
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