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A company assigns new interns their cabins in such a way that some of
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01 Sep 2021, 09:25
01 Sep 2021, 09:25
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Question Stats:
66% (01:23) correct
33% (00:56) wrong based on 33 sessions
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A company assigns new interns their cabins in such a way that some of the cabins can be left empty and more than one intern can be assigned to one cabin.
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Number of ways in which 3 interns can be assigned to 2 different cabins |
9 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Re: A company assigns new interns their cabins in such a way that some of
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01 Sep 2021, 10:09
01 Sep 2021, 10:09
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Is it 3C1 * 3C1 ?
A company assigns new interns their cabins in such a way that some of
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01 Sep 2021, 18:15
01 Sep 2021, 18:15
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This is an interesting one. It can be interpreted two ways:
Assuming that we must assign all the interns to at least one of the two cabins, the answer is B.
Assuming that the interns do not need to be assigned to any of the cabins (we don't assign both cabins any interns), the answer is A.
I'm going to assume the first assumption and can clarify why the answer would be A if we're using the second assumption.
We have Cabin \(A\) and Cabin \(B\).
If I assign 3 interns to cabin \(A\), then none are assigned to Cabin \(B\) \((3,0)\)
If I assign 2 interns to cabin \(A\), then 1 is assigned to Cabin \(B\) \((2,1)\)
If I assign 1 intern to cabin \(A\), then 2 are assigned to Cabin \(B\) \((1,2)\)
If I assign 0 interns to cabin \(A\), then 3 are assigned to Cabin \(B\) \((0,3)\)
In the case where we assign 2 to one cabin and 1 to the other, we can look at the different combinations of the interns in each.
So we have \(3C2\) (We have 3 interns, assign 2 to one cabin). Do this twice because we have two cases (\((2,1)\) and \((1,2)\)).
\(3C2 = 3\), and since we have it twice, there 6 ways to arrange the interns this way.
There is only one way to assign three interns to one cabin, and we have two cases of this too (\((3,0)\) and \((0,3)\)). So that gives us 2 more arrangements.
Summing the different cases, we have \(1+1+3+3 = 8\).
So the answer is B
_________________
If we used the second assumption, then we can have a case where we assign no intern to any cabin \((0,0)\), or only one to a cabin, and the rest don't get assigned a cabin \((0,1)\), etc.
It's clear that with this assumption, we would have more than 9 arrangements. So the answer would be A. However, I would be surprised if this was the case.
Assuming that we must assign all the interns to at least one of the two cabins, the answer is B.
Assuming that the interns do not need to be assigned to any of the cabins (we don't assign both cabins any interns), the answer is A.
I'm going to assume the first assumption and can clarify why the answer would be A if we're using the second assumption.
We have Cabin \(A\) and Cabin \(B\).
If I assign 3 interns to cabin \(A\), then none are assigned to Cabin \(B\) \((3,0)\)
If I assign 2 interns to cabin \(A\), then 1 is assigned to Cabin \(B\) \((2,1)\)
If I assign 1 intern to cabin \(A\), then 2 are assigned to Cabin \(B\) \((1,2)\)
If I assign 0 interns to cabin \(A\), then 3 are assigned to Cabin \(B\) \((0,3)\)
In the case where we assign 2 to one cabin and 1 to the other, we can look at the different combinations of the interns in each.
So we have \(3C2\) (We have 3 interns, assign 2 to one cabin). Do this twice because we have two cases (\((2,1)\) and \((1,2)\)).
\(3C2 = 3\), and since we have it twice, there 6 ways to arrange the interns this way.
There is only one way to assign three interns to one cabin, and we have two cases of this too (\((3,0)\) and \((0,3)\)). So that gives us 2 more arrangements.
Summing the different cases, we have \(1+1+3+3 = 8\).
So the answer is B
_________________
If we used the second assumption, then we can have a case where we assign no intern to any cabin \((0,0)\), or only one to a cabin, and the rest don't get assigned a cabin \((0,1)\), etc.
It's clear that with this assumption, we would have more than 9 arrangements. So the answer would be A. However, I would be surprised if this was the case.
