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Rasalghul853
Joined: 24 Jan 2019
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Updated on: 12 Apr 2021, 08:08
Originally posted by Rasalghul853 on 12 Apr 2021, 06:07.
Last edited by Bunuel on 12 Apr 2021, 08:08, edited 1 time in total.
Last edited by Bunuel on 12 Apr 2021, 08:08, edited 1 time in total.
Renamed the topic and added the OA.
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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)!
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73% (01:35) correct
27%
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From among a couple of 6, a committee of 5 has to be selected. If the committee has no couple how many ways can the committee be formed?
A. 32
B. 192
C. 576
D. 720
E. 792
A. 32
B. 192
C. 576
D. 720
E. 792
12 Apr 2021, 08:11
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Rasalghul853
Each couple can send only one "representative" to the committee. We can choose 5 couples (as there should be 5 members) to send only one "representatives" to the committee in \(C^5_6=6\) # of ways.
But these 5 chosen couples can send one of two persons (either husband or wife): \(2*2*2*2*2=2^5\).
So, the number of ways to choose 5 people out 6 married couples so that none of them would be married to each other is: \(C^5_6*2^5=192\).
Answer: B.
12 Apr 2021, 08:56
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Hello, everyone. First off, I think the question could use a little rephrasing:
FROM: From among a couple of 6... If the committee has no couple...
TO: From among 6 couples... If the committee has/consists of no couples...
The former phrasing allows us in the first part to conjure up a polyamorous "couple" whose entourage consists of four other individuals, since a couple is a singular entity; the latter leaves little room for interpretation, assuming that couples consist of two individuals, as is often the case in such questions. That said, I agree with the two solutions posted above, but to add something more constructive to the dialogue, I thought I would add a less formulaic, more logic-based way of arriving at the correct answer, in case any onlookers find permutations, combinations, and combinatorics a bit too much to handle.
An alternative method: Think of the committee consisting of 5 slots to fill: __ __ __ __ __
The number of ways all these people could be selected up to this point is
\(12*10*8*6*4\)
I would not recommend multiplying this out to derive the product. Why? Because we need to divide out the repetitions of selecting a given person in each of the five slots. That is, of the five people selected for the committee, any one of them could have been selected first, but once selected, that same person cannot be chosen again. We can thus create a new set of five slots and work with this smaller group in a similar manner as we had above: __ __ __ __ __
So, the following would show the number of ways to arrange these 5 committee members:
\(5*4*3*2*1\)
For the final step, we can take the product from before and divide by our "committee" product above:
\(\frac{12*10*8*6*4}{5*4*3*2*1}\)
Since we left each product alone, we can now reduce much easier. The 4s cancel out, and 6 can also be reduced to 1 in the numerator. We can make quick work of the rest:
\(\frac{12*10*8}{5}\)
\(\frac{960}{5}\) or
\(\frac{12*2*8}{1}\)
\(192\)
The answer must be (B).
This approach may not be as fast as those outlined in earlier posts, but it can help someone who might forget such a method on the spot. (Just for fun, I used my alternative approach, and the question took little more than a minute.)
Happy studies, everyone.
- Andrew
FROM: From among a couple of 6... If the committee has no couple...
TO: From among 6 couples... If the committee has/consists of no couples...
The former phrasing allows us in the first part to conjure up a polyamorous "couple" whose entourage consists of four other individuals, since a couple is a singular entity; the latter leaves little room for interpretation, assuming that couples consist of two individuals, as is often the case in such questions. That said, I agree with the two solutions posted above, but to add something more constructive to the dialogue, I thought I would add a less formulaic, more logic-based way of arriving at the correct answer, in case any onlookers find permutations, combinations, and combinatorics a bit too much to handle.
An alternative method: Think of the committee consisting of 5 slots to fill: __ __ __ __ __
- There are any of 12 ways to fill that first slot, since there are 12 people to choose from.
- There are 10 ways to fill the second slot, since we have to remove the first person who was selected (in slot one) and that person's partner.
- There are 8 ways to fill the third slot, repeating the process from the last step—i.e. we cannot select either the two previously selected people or their partners.
- There are 6 ways to fill the fourth slot. (Notice a pattern here?)
- There are 4 ways to fill the last slot from the remaining individuals.
The number of ways all these people could be selected up to this point is
\(12*10*8*6*4\)
I would not recommend multiplying this out to derive the product. Why? Because we need to divide out the repetitions of selecting a given person in each of the five slots. That is, of the five people selected for the committee, any one of them could have been selected first, but once selected, that same person cannot be chosen again. We can thus create a new set of five slots and work with this smaller group in a similar manner as we had above: __ __ __ __ __
- There are 5 ways to fill the first slot, since 5 people were selected for the committee.
- There are 4 ways to fill the second slot, since one person would already have been selected.
- There are 3 ways to fill the third slot, repeating the process from the last step.
- There are 2 ways to fill the penultimate slot.
- There is a single way to fill the final slot, since everyone else would have been selected already.
So, the following would show the number of ways to arrange these 5 committee members:
\(5*4*3*2*1\)
For the final step, we can take the product from before and divide by our "committee" product above:
\(\frac{12*10*8*6*4}{5*4*3*2*1}\)
Since we left each product alone, we can now reduce much easier. The 4s cancel out, and 6 can also be reduced to 1 in the numerator. We can make quick work of the rest:
\(\frac{12*10*8}{5}\)
\(\frac{960}{5}\) or
\(\frac{12*2*8}{1}\)
\(192\)
The answer must be (B).
This approach may not be as fast as those outlined in earlier posts, but it can help someone who might forget such a method on the spot. (Just for fun, I used my alternative approach, and the question took little more than a minute.)
Happy studies, everyone.
- Andrew
