Problem Solving

vikram4689 wrote:Hi Mitch,

newgmattest's approach seems to be similar to your approach. But the answers are different. can you please explain this

Hi,
newgmattest has multiplied the end result with 2, which is incorrect. By doing so, he is considering AB and BA as different pairs. That is the reason he won't be getting 4C2 pairs, he will be getting 4P2 pairs.

vikram4689 wrote:Hi Mitch,

newgmattest's approach seems to be similar to your approach. But the answers are different. can you please explain this

Newgmattest's approach:

first writer can get the job is C(4,2).
The no. of ways the second writer can get one of the remaining two jobs is C(2,1).
There is only one way the third writer can get the remaining job.

So your numerator is C(4,2)* C(2,1)*1

4C2 = 6 = how many ways a pair of tasks can be assigned to the first writer.
2! = 2*1 = 2 = how many ways the remaining 2 tasks can be assigned to the other 2 writers.
Multiplying these results, we get:
6*2 = 12 possible assignments.

Since any of the 3 writers can be assigned the pair of tasks, the result above must be multiplied by 3:
Number of ways in which each writer is assigned at least one task = 3*12 = 36.

No reason to multiply by 2.

Whenever we finish calculating an answer, we should always confirm that the result makes sense.
8/9 is much too high a probability.
For each writer to be assigned at least one task, the tasks must be divided 2-1-1.
Distributions in which each writer is NOT assigned at least one task are 4-0-0, 3-1-0, 2-2-0.
Since there are more ways in which a writer could NOT be assigned at least one task, P(each writer is assigned at least one task) = 8/9 is impossible.

Thanks Mitch, Now i am trying to find out where i went wrong.I calculated in the following manner

Options for companies to get writers task = 3*2*1*3

I think i somehow missed the 1st of where you got 6 and i am getting 3.

GMATGuruNY wrote:4C2 = 6 = how many ways a pair of tasks can be assigned to the first writer.

For the problem ahead, we were on the same path as in you chose 2*1 for remaining 2 writers and multiply with 3 as any one of the writers can be assigned the 3rd job.

I choose 3 ways because job 1 can be assigned to writer 1 in 3 ways. I did not consider a pair because i thought it is not necessary as i am multiplying with 3 to consider the effect of the 4th job.

Can you please highlight my mistake.

vikram4689 wrote:

Options for companies to get writers task = 3*2*1*3

I think i somehow missed the 1st of where you got 6 and i am getting 3.

Hi,
1st company has 3 choices.
You are saying second company has 2 choices. That essentially means 2nd company cannot take the worker who was picked by company 1, which is not the case.Similarly, you are saying that 3rd company cannot take either of workers picked by company 1 or company 2.
Let the workers be a,b,c and we write the workers picked by companies 1,2,3,4 in order
I will write the cases that are missing from your count
case-1: 2nd company can pick the same guy picked by 1st company
Then number of ways will be 3*1*2*1
Example: a,a,b,c
a,a,c,b
b,b,a,c
b,b,c,a
c,c,a,b
c,c,b,a

case-2: 3rd company picks worker of 1st company
a,b,a,c
a,c,a,b
b,a,b,c
b,c,b,a
c,a,c,b
c,b,c,a

case-3 3rd company picks worker of 2nd comapny
Here also you will get6 combinations.

So, add this 18 to your 3*2*1*3(this is the case when companies 1,2,3 take different workers but it is not necessarily true)
The new sum will be 18+3*2*1*3 = 36.

I have a basic doubt here. 4 companies have 4 tasks that are to be given to 3 different writers.

Writer 1 can get one task , or 2 tasks or 3 tasks or 4 tasks.
So he can get tasks in 4 ways.
the same with other writers.
so can't the total events be 4*4*4 i.e. 64?
I'm little confused

baladon99 wrote:I have a basic doubt here. 4 companies have 4 tasks that are to be given to 3 different writers.

Writer 1 can get one task , or 2 tasks or 3 tasks or 4 tasks.
So he can get tasks in 4 ways.
the same with other writers.
so can't the total events be 4*4*4 i.e. 64?
I'm little confused

Hi,
The underlying assumption is that each task can be performed by a single writer at a given time, although the wording of the problem is bad in this case. So, you are counting the case of same task being performed by all 3 writers at the same time, which is not the case.
So, each task has three options for writers. So, it will be 3^4 and not 4^3

Got it , All of these process are a bit convoluted, wonder how i would be able to solve on test day.