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In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory....- May 05
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06 Apr 2020, 04:13
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VeritasKarishma
Quote:
The only error here is highlighted. You do not pick one of the two secretaries.
Say you have 3 Secys - S1, S2, S3
When you pick 3C1 and then 2C1, you are arranging them as first and second. But you do not need to do that.
Just 3C1 (or 3C2 which is the same) is enough. It gives you 2 secys.
Say you get S1 and S2. Each will get 2 reports. When you choose 2 of the 4 reports to give to S1, say you choose R1 and R2 to give to S1, it gives all cases in which the reports can given to the 2 secys. This is so because there will be a case when R3 and R4 are given to S1 and R1 and R2 are given to S2.
27 Jul 2020, 00:52
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This is an interesting Distribution question. It took me around 3 minutes to think about the right approach. I think what VeritasKarishma has described is a straight forward way of calculating the chances where you subtract the ways at least one secretary doesn't get any letter from all the possible cases.
I thought a different way. You need to think assigning secretaries to the department.
So, Dep1 can have 3 ways to assign a secretary
Dep2 can have 3 ways to assign a secretary
Dep3 can have 3 ways to assign a secretary
Dep4 can have 3 ways to assign a secretary
Now to give at least one department to each secretary means that assign 3 departments to 3 secretaries => 4C3 ways
For the last department, there are 3 ways to assign it to the secretaries.
Lastly, you cannot assign the 4th department to a fourth secretary, hence total cases will not be 3*3*3*3, but 3*3*3 and the last department can be assigned to any secretary.
So total desirable = 4C3 * 3 / (3*3*3)
= 4/ 9
I thought a different way. You need to think assigning secretaries to the department.
So, Dep1 can have 3 ways to assign a secretary
Dep2 can have 3 ways to assign a secretary
Dep3 can have 3 ways to assign a secretary
Dep4 can have 3 ways to assign a secretary
Now to give at least one department to each secretary means that assign 3 departments to 3 secretaries => 4C3 ways
For the last department, there are 3 ways to assign it to the secretaries.
Lastly, you cannot assign the 4th department to a fourth secretary, hence total cases will not be 3*3*3*3, but 3*3*3 and the last department can be assigned to any secretary.
So total desirable = 4C3 * 3 / (3*3*3)
= 4/ 9
05 Aug 2020, 05:41
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VeritasKarishma
Hi VeritasKarishma, so, we have 4 reports and 3 secretaries. The question is: in how many ways can we give 1 report to 1 secretary and 3 reports to the other secretary?
Why can the solution not be the following:
1 report to 1 secretary in 4C1*3C1 ways
3 reports to the other secretary in 2C1 ways (2C1 because there are two secretaries left and anyone of them will get the remaining 3 reports).
So, ways in which we can give 1 report to 1 secretary and 3 reports to the other secretary = 4C1*3C1*2C1 = 24.
But apparently this is not correct. Where am I going wrong?
