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05 Jan 2014, 19:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
42% (02:19) correct
58%
(02:36)
wrong
based on 119
sessions
History
Date
Time
Result
Not Attempted Yet
The number of ways a lawn tennis mixed doubles match can be made up of 7 married couples if no husband and wife plays in the same match:
A. 210
B. 420
C. 840
D. 920
E. 1050
A. 210
B. 420
C. 840
D. 920
E. 1050
I have a couple of questions on P n C and it will be great if someone can help me out with the concept/ process of solving.
In P n C, the questions involving pairs always knocks me out and I guess its because of lack of concept. So some additional problems/approach can help me tune up my skills in this area, and your help is highly appreciated.
Below are the questions:
1) The number of ways a lawn tennis mixed doubles match can be made up of 7 married couples if no husband and wife plays in the same match:
a) 210 b) 420 c) 840 d) None
Assumed correct answer is 420 and here is how it has been reached:
Choose two husbands out of 7: 7C2 = 21
Choose two wives : 5C2 = 10
No. of teams = 21 x 10 = 210.
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 210 x 2 =420.
Pretty easy to understand and I have no problem accepting it. However, the problem arises when I go by pairs and here is my question to you why the below solution is not working out:
Team A:
Select a husband or wife from 7 pairs = 7 x 2 = 14
Select the other spouse from 6 pairs = 6
Team B: Select a husband or wife from 5 pairs = 5 x 2 = 10
Select the other spouse from 4 pairs = 4
Total selections = 14 x 6 x 4 x 10 = 84 x 40 = 3360
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 3360 x 2 = 6720
Can someone explain why the above approach is incorrect?
In P n C, the questions involving pairs always knocks me out and I guess its because of lack of concept. So some additional problems/approach can help me tune up my skills in this area, and your help is highly appreciated.
Below are the questions:
1) The number of ways a lawn tennis mixed doubles match can be made up of 7 married couples if no husband and wife plays in the same match:
a) 210 b) 420 c) 840 d) None
Assumed correct answer is 420 and here is how it has been reached:
Choose two husbands out of 7: 7C2 = 21
Choose two wives : 5C2 = 10
No. of teams = 21 x 10 = 210.
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 210 x 2 =420.
Pretty easy to understand and I have no problem accepting it. However, the problem arises when I go by pairs and here is my question to you why the below solution is not working out:
Team A:
Select a husband or wife from 7 pairs = 7 x 2 = 14
Select the other spouse from 6 pairs = 6
Team B: Select a husband or wife from 5 pairs = 5 x 2 = 10
Select the other spouse from 4 pairs = 4
Total selections = 14 x 6 x 4 x 10 = 84 x 40 = 3360
No wives and husbands in the two teams can be swapped considering it is mixed doubles: 3360 x 2 = 6720
Can someone explain why the above approach is incorrect?
06 Jan 2014, 17:21
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AGH
Dear AGH
I'm happy to respond.
First of all, here's a blog post I think you will find helpful:
https://magoosh.com/gmat/2013/difficult- ... -problems/
One of the hardest things about counting question is to recognize when you are counting redundancies --- that is, when you are using a counting method that allows for the same combination to be counted in multiple ways. If your calculation produces a number much bigger than the OA, then it's always the case that you are are counting redundancies --- i.e., the same situation gets counted more than once.
Let's say the seven husbands are {a, b, c, d, e, f, g} and the seven corresponding wives are {A, B, C, D, E, F, G}.
Now, consider your method:
first choice = one of fourteen people
second choice = one of six non-spouses of first choice
third choice = one of the ten people in unused couples
fourth choice = one of four remaining non spouses
Consider the mixed doubles match that results from the four selections: {e, B, G, d}
Those same two teams would also be created by the selections:
{B, e, G, d}, {e, B, d, G}, {B, e, d, G}, {G, d, e, B}, {d, G, e, B}, {G, d, B, e}, and {d, G, B, e}.
Right away, that shows eight different ways that the same two teams can be selected using your approach, which means your answer of 3360 is 8 times too big. 3360/8 = 420, because it already would include all the swapped teams as separate choices.
Once again, counting problems are tricky: you have to be extremely careful that you are not counting the same thing more than once. The perspective you take in framing the counting problem is crucial.
Does all this make sense?
Mike
General Discussion
15 May 2025, 12:55
Kudos
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Here is the step-by-step solution to find the number of ways to form a mixed doubles team from 7 married couples such that no husband and wife play together.
Step 1: Select 2 men
We need to select 2 men from the 7 husbands. The number of ways to do this is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\).
Number of ways to select 2 men = \(C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21\).
Step 2: Select 2 women such that their husbands were not selected
Since the selected men cannot play with their wives, the two women selected must be from the remaining 5 wives whose husbands were not chosen.
Number of ways to select 2 women = \(C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10\).
Step 3: Form the two mixed teams
We have selected 2 men and 2 women. Let the selected men be M1 and M2, and the selected women be W1 and W2. We need to form two mixed double teams. There are two possible pairings:
Step 4: Calculate the total number of ways
To find the total number of ways to form the two mixed doubles teams such that no husband and wife play together, we multiply the number of ways to perform each step.
Total number of ways = (Number of ways to select 2 men) × (Number of ways to select 2 women) × (Number of ways to form teams)
Total number of ways = \(21 \times 10 \times 2 = 420\)
Final Answer
The number of ways a lawn tennis mixed doubles match can be made up of 7 married couples if no husband and wife plays in the same match is \(420\).
Key Concept & Explanation
The key concept used here is combinations, which is a method of selecting items from a set where the order of selection does not matter. We used combinations to select the men and women for the teams. The constraint that no husband and wife play together affects the selection of the women based on the selection of men. We then considered the number of ways to pair the selected men and women into two teams.
Step 1: Select 2 men
We need to select 2 men from the 7 husbands. The number of ways to do this is given by the combination formula \(C(n, k) = \frac{n!}{k!(n-k)!}\).
Number of ways to select 2 men = \(C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} = \frac{7 \times 6}{2 \times 1} = 21\).
Step 2: Select 2 women such that their husbands were not selected
Since the selected men cannot play with their wives, the two women selected must be from the remaining 5 wives whose husbands were not chosen.
Number of ways to select 2 women = \(C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10\).
Step 3: Form the two mixed teams
We have selected 2 men and 2 women. Let the selected men be M1 and M2, and the selected women be W1 and W2. We need to form two mixed double teams. There are two possible pairings:
- Team 1: M1 and W1, Team 2: M2 and W2
- Team 1: M1 and W2, Team 2: M2 and W1
Step 4: Calculate the total number of ways
To find the total number of ways to form the two mixed doubles teams such that no husband and wife play together, we multiply the number of ways to perform each step.
Total number of ways = (Number of ways to select 2 men) × (Number of ways to select 2 women) × (Number of ways to form teams)
Total number of ways = \(21 \times 10 \times 2 = 420\)
Final Answer
The number of ways a lawn tennis mixed doubles match can be made up of 7 married couples if no husband and wife plays in the same match is \(420\).
Key Concept & Explanation
The key concept used here is combinations, which is a method of selecting items from a set where the order of selection does not matter. We used combinations to select the men and women for the teams. The constraint that no husband and wife play together affects the selection of the women based on the selection of men. We then considered the number of ways to pair the selected men and women into two teams.
17 May 2025, 06:36
Kudos
Bookmarks
A male can be chosen 7 ways.
His woman playing partner can be chosen 6 ways since his wife is disallowed, so:
7*6= 42 ways
The male on the other team can be chosen 5 ways since the husband of the woman chosen above is disallowed.
His woman playing partner can be chosen 4 ways since his and the other man's wife are disallowed and 1 woman has already been chosen, so:
5*4= 20 ways
So, preliminary number of ways is:
42*20 = 840
Since the above process unnecessarily distinguishes between an ab cd match and a cd ab match (it doesn't matter which side of the net they start playing 😀) , the above result should be divided by 2:
1/2 * 840 = 420
His woman playing partner can be chosen 6 ways since his wife is disallowed, so:
7*6= 42 ways
The male on the other team can be chosen 5 ways since the husband of the woman chosen above is disallowed.
His woman playing partner can be chosen 4 ways since his and the other man's wife are disallowed and 1 woman has already been chosen, so:
5*4= 20 ways
So, preliminary number of ways is:
42*20 = 840
Since the above process unnecessarily distinguishes between an ab cd match and a cd ab match (it doesn't matter which side of the net they start playing 😀) , the above result should be divided by 2:
1/2 * 840 = 420
