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01 Dec 2018, 02:31
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vishuvashishth
Sure, I will help you out on any questions you might have.
Note that not every question given by you will involve making equations. But I will go one line at a time and show you how to evaluate it to arrive at the answer.
Here is the first one: https://gmatclub.com/forum/of-the-stude ... l#p2182147
01 Dec 2018, 02:49
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vishuvashishth
No. 3
https://gmatclub.com/forum/in-order-to- ... l#p2182154
This doesn't involve forming equations either. In fact, there are very few questions in which you will need to actually form an equation and then solve it. Notice the way you should evaluate this sentence to get to the answer.
01 Dec 2018, 02:49
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Question 2: A group of 10 people consists of 2 married couples and 6 bachelors. A committee of 4 is to be selected from the 10 people. How many different committees can be formed if the committee can consist of at most one married couple?
Solution: We have to select 4 people out of: 6 bachelors and 2 married couples.
The number of ways of selecting any 4 people out of 10 is 10*9*8*7/4! = 210 (Note here that we are just selecting 4 people. We are not arranging them so we divide by 4!)
The people will get selected in various ways:
1. Four bachelors
2. One from a couple and three bachelors
3. Two from two different couples and two bachelors
4. One couple and two bachelors
5. One couple, one person from a couple, one bachelor
6. Two couples
If we add the number of committees possible in each of these cases, we will get 210. Out of all these cases, only the last one (two couples) has more than one married couple. Instead of calculating the number of different committees that can be formed in each of the first five cases, we can calculate the number of committees in the last case and subtract it from 210.
How many different committees can be formed such that there are 2 couples? Only one since we have only 2 couples. We will have to select both the couples and we will get 4 people.
Number of different committees of 4 people such that there is at most one married couple = 210 – 1 = 209.
Just for practice, let’s see how we can calculate the different number of committees that can be formed in each of the first five cases. The sum of all these cases should give us 209.
1. Select 4 bachelors from 6 bachelors in 6*5*4*3/4! = 15 different committees
2. Select 1 person out of the two couples (4 people) in 4 ways and 3 bachelors from 6 bachelors in 6*5*4/3! = 20 ways. So you select the 4 people in 4*20 = 80 different committees
3. Select 2 people from 2 different couples in 4*2/2! = 4 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 4*15 = 60 different committees
4. Select 1 couple in 2 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 2*15 = 30 different committees
5. Select 1 couple in 2 ways, 1 person from the remaining couple in 2 ways and 1 bachelor from 6 bachelors in 6 ways. So you can select the 4 people in 2*2*6 = 24 different committees
The sum of all these five cases = 15 + 80 + 60 + 30 + 24 = 209 different committees
Hi VeritasKarishma
First of all,Thanks for making these articles.
Now I have a slight doubt related to above question .
4. Select 1 couple in 2 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 2*15 = 30 different committees
5. Select 1 couple in 2 ways, 1 person from the remaining couple in 2 ways and 1 bachelor from 6 bachelors in 6 ways. So you can select the 4 people in 2*2*6 = 24 different committees
When I was doing the question, I considered 5 and 6 to be symmetrical cases.
I calculated it as follows:
2c1(1 married couple selected) +
2c1(One member of the other married couple selected) +
7c2(selected 2 members out of a group of 7)
= 84 ways
Now i do realise that I am counting something twice. But I can't exactly point out where.
Can you please tell me where I am going wrong, and If I were to proceed with my method, what should I have subtracted to get a correct answer.
Hope to hear from you soon!
Regards
Nitesh
Solution: We have to select 4 people out of: 6 bachelors and 2 married couples.
The number of ways of selecting any 4 people out of 10 is 10*9*8*7/4! = 210 (Note here that we are just selecting 4 people. We are not arranging them so we divide by 4!)
The people will get selected in various ways:
1. Four bachelors
2. One from a couple and three bachelors
3. Two from two different couples and two bachelors
4. One couple and two bachelors
5. One couple, one person from a couple, one bachelor
6. Two couples
If we add the number of committees possible in each of these cases, we will get 210. Out of all these cases, only the last one (two couples) has more than one married couple. Instead of calculating the number of different committees that can be formed in each of the first five cases, we can calculate the number of committees in the last case and subtract it from 210.
How many different committees can be formed such that there are 2 couples? Only one since we have only 2 couples. We will have to select both the couples and we will get 4 people.
Number of different committees of 4 people such that there is at most one married couple = 210 – 1 = 209.
Just for practice, let’s see how we can calculate the different number of committees that can be formed in each of the first five cases. The sum of all these cases should give us 209.
1. Select 4 bachelors from 6 bachelors in 6*5*4*3/4! = 15 different committees
2. Select 1 person out of the two couples (4 people) in 4 ways and 3 bachelors from 6 bachelors in 6*5*4/3! = 20 ways. So you select the 4 people in 4*20 = 80 different committees
3. Select 2 people from 2 different couples in 4*2/2! = 4 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 4*15 = 60 different committees
4. Select 1 couple in 2 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 2*15 = 30 different committees
5. Select 1 couple in 2 ways, 1 person from the remaining couple in 2 ways and 1 bachelor from 6 bachelors in 6 ways. So you can select the 4 people in 2*2*6 = 24 different committees
The sum of all these five cases = 15 + 80 + 60 + 30 + 24 = 209 different committees
Hi VeritasKarishma
First of all,Thanks for making these articles.
Now I have a slight doubt related to above question .
4. Select 1 couple in 2 ways and 2 bachelors from 6 bachelors in 6*5/2! = 15 ways. So you select the 4 people in 2*15 = 30 different committees
5. Select 1 couple in 2 ways, 1 person from the remaining couple in 2 ways and 1 bachelor from 6 bachelors in 6 ways. So you can select the 4 people in 2*2*6 = 24 different committees
When I was doing the question, I considered 5 and 6 to be symmetrical cases.
I calculated it as follows:
2c1(1 married couple selected) +
2c1(One member of the other married couple selected) +
7c2(selected 2 members out of a group of 7)
= 84 ways
Now i do realise that I am counting something twice. But I can't exactly point out where.
Can you please tell me where I am going wrong, and If I were to proceed with my method, what should I have subtracted to get a correct answer.
Hope to hear from you soon!
Regards
Nitesh
01 Dec 2018, 02:53
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vishuvashishth
I have already solved no 2 by the variation method. I have also provided the link to my relevant post there.
You can also solve it using unitary method as shown by some people there (if 10 workers make 50 chairs, 1 worker will make 5 chairs so 18 workers will make 18*5 chairs etc). But I prefer the simple and straight forward method shown by me. Once you understand it, it takes a few seconds to implement.
