5 couples say a b c d e
2 2 2 2 2
Now if we apply the condition then we can select one from each in 5C3 ways ..so i think it shd be 5c3 * 2 = 20
Please correct me if i am wrong
help =)
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sujaysolanki
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1. Selecting the first member - There are 10 ways of selecting the first person out of a total of 10 people.
2. Selecting the second member - Now you are left with only 8 members to select from because the spouse of the already selected member is not eligible.
3. Selecting the last member - Again the spouse of the 2nd member is not eligible so you are left with only 6 people to choose from.
So the total number of ways the committee can be formed is:
10*8*6 = 480.
However, the three members within each group can be arranged in 3! ways (meaning there are 6 occurences of the same group in the total).
Hence the total number of distinct groups that you can select is 480/6 = 80.
Regards,
Samsonite
2. Selecting the second member - Now you are left with only 8 members to select from because the spouse of the already selected member is not eligible.
3. Selecting the last member - Again the spouse of the 2nd member is not eligible so you are left with only 6 people to choose from.
So the total number of ways the committee can be formed is:
10*8*6 = 480.
However, the three members within each group can be arranged in 3! ways (meaning there are 6 occurences of the same group in the total).
Hence the total number of distinct groups that you can select is 480/6 = 80.
Regards,
Samsonite
Number of ways of picking the first group = 5
Number of ways of picking a member from this group = 2
Number of ways of picking the second group = 4
Number of ways of picking a member from this group = 2
Number of ways of picking the last group = 3
Number of ways of picking a member from this group = 2
Total number of arrangements = 5*2*4*2*3*2 = 480
Number of redundant groups (because H1W2H3 is the same as W2H1H3) = 3! = 6.
Hence the number of ways of picking 3 members is 480/6 = 80.
Remember: Combination = Permutation/redundancy
Regards,
Samsonite
Number of ways of picking a member from this group = 2
Number of ways of picking the second group = 4
Number of ways of picking a member from this group = 2
Number of ways of picking the last group = 3
Number of ways of picking a member from this group = 2
Total number of arrangements = 5*2*4*2*3*2 = 480
Number of redundant groups (because H1W2H3 is the same as W2H1H3) = 3! = 6.
Hence the number of ways of picking 3 members is 480/6 = 80.
Remember: Combination = Permutation/redundancy
Regards,
Samsonite
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sujaysolanki
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Orwell_Jetski
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