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30 Mar 2020, 04:22
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Five married couple, how many ways 3 member team can you form such that the couple will not be together?
A. 75
B. 80
C. 120
D. 90
E. 85
A. 75
B. 80
C. 120
D. 90
E. 85
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30 Mar 2020, 05:14
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I think the answer should be 80.
We have 5 couples, i.e. 10 people.
Now, for the first slot, we have 10 choices, for the second slot, we have 8 choices (one person has already filled slot 1 and his/her partner can no more be chosen as per the condition in the question), and for the third slot, we have 6 choices.
So, 10*8*6 = number of ways we can choose a three-member team but since the teams are not different basis the order of the slot filled by the person, we divide this by 3!.
So, 10*8*6/3/2 = 80.
Was 80 the correct answer?
We have 5 couples, i.e. 10 people.
Now, for the first slot, we have 10 choices, for the second slot, we have 8 choices (one person has already filled slot 1 and his/her partner can no more be chosen as per the condition in the question), and for the third slot, we have 6 choices.
So, 10*8*6 = number of ways we can choose a three-member team but since the teams are not different basis the order of the slot filled by the person, we divide this by 3!.
So, 10*8*6/3/2 = 80.
Was 80 the correct answer?
31 Mar 2020, 00:33
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Hello Bijay,
This is a question where you can use the reverse engineering approach of solving combinations questions. Not sure what I’m telling? Remember that in one of my sessions on Combinatorics, I had given out a principle,
What I want = Total – What I don’t want.
So, what do we not want here? We do not want the team to have a couple.
IN how many ways can a couple be a part of a team?? We need a TEAM OF 3 persons. Of these 3 persons, if we have ONE COUPLE, we already have 2 persons, isn’t it? How many MORE will we have to select? We will need one more person, apart from the couple we selected.
We have 5 couples, so selecting ANY ONE of them can be done in \(5_C_1\) ways.
Once this is done, we will be left with 8 persons. Of the 8 persons, we need to select ANY ONE; this can be done in \(8_C_1\) ways,.
Therefore, the number of ways of selecting a team of 3 persons which HAS ONE COUPLE = \(5_C_1\) * \(8_C_1\) = 5 * 8 = 40.
Do we want the above 40? No, we don’t. Therefore, we subtract this from the TOTAL number of 3-member teams that can be formed.
Total number of 3-member teams that can be formed = \(10_C_3\) = 120.
Therefore, Number of 3-member teams WITHOUT a couple = 120 – 40 = 80.
The correct answer option is B, IMO.
Hope that helps!
This is a question where you can use the reverse engineering approach of solving combinations questions. Not sure what I’m telling? Remember that in one of my sessions on Combinatorics, I had given out a principle,
What I want = Total – What I don’t want.
So, what do we not want here? We do not want the team to have a couple.
IN how many ways can a couple be a part of a team?? We need a TEAM OF 3 persons. Of these 3 persons, if we have ONE COUPLE, we already have 2 persons, isn’t it? How many MORE will we have to select? We will need one more person, apart from the couple we selected.
We have 5 couples, so selecting ANY ONE of them can be done in \(5_C_1\) ways.
Once this is done, we will be left with 8 persons. Of the 8 persons, we need to select ANY ONE; this can be done in \(8_C_1\) ways,.
Therefore, the number of ways of selecting a team of 3 persons which HAS ONE COUPLE = \(5_C_1\) * \(8_C_1\) = 5 * 8 = 40.
Do we want the above 40? No, we don’t. Therefore, we subtract this from the TOTAL number of 3-member teams that can be formed.
Total number of 3-member teams that can be formed = \(10_C_3\) = 120.
Therefore, Number of 3-member teams WITHOUT a couple = 120 – 40 = 80.
The correct answer option is B, IMO.
Hope that helps!
