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38 votes
38 votes

There are $6$ jobs with distinct difficulty levels, and $3$ computers with distinct processing speeds. Each job is assigned to a computer such that:

  • The fastest computer gets the toughest job and the slowest computer gets the easiest job.
  • Every computer gets at least one job.

The number of ways in which this can be done is ___________.

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3 Answers

Best answer
62 votes
62 votes
Let $C_1$ be the fastest and $C_3$ be the slowest computers.

These two are assigned two jobs. Now out of the remaining $4$ jobs we need to ensure $C_2$ gets at least $1.$ Without this constraint we can assign $4$ jobs to $3$ computers in $3^4=81$ ways. Out of these $81$ ways $2^4=16$ will be having no jobs for $C_2.$

So, number of possible ways so that $C_2$ gets at least one job $=81-16=65.$
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44 votes
44 votes

First let me assume that the computers are $C_1,C_2,C_3$ having computing power in the increasing order as shown below:

And let me assume that the jobs are $J_1,J_2,J_3,J_4,J_5.J_6$ with increasing difficulty level as shown below:

Now as per the condition “The fastest computer gets the toughest job and the slowest computer gets the easiest job.” the situation is as follows:

Which means that $J_1$ is assigned to $C_1$ and $J_6$ is assigned to $C_3$.

Now note the second condition: “Every computer gets at least one job.”

Based on the above condition, at least one of the jobs $J_2,J_3.J_4,J_5$ must be assigned to $C_2$, or else $C_2$ computer shall remain vacant!!!

Now to calculate the number of ways in which at least one of $J_2,J_3.J_4,J_5$ is assigned to $C_2$, we find the all ways in which the above $4$ jobs can be assigned to computers and from that we subtract the cases in which no job is assigned to $C_2$.

The total number of possible ways= $3^4 =81$ [As each of the above $4$ jobs in the yellow bubble can be assigned to either computer $C_1,C_2,C_3$]

The number of ways in which in the above $4$ jobs are assigned only to $C_1$ or $C_3$=$2^4$ =16[as each of the jobs has $2$ choices]

So required number=$81-16=65$

4 votes
4 votes

According to given conditions, we have to find number of onto function with certain condition (i.e., The fastest computer gets the toughest job and the slowest computer gets the easiest job.),

Therefore, we have remaining 4 jobs to assign these 3 computer such that Every computer gets at least one job.

Since, fasted and slowest computer has already atleast one-one job, we assigned.
Only Computer with medium speed has no job yet, so we need to remove combination of with no job in Computer with medium speed.

Therefore,
= $3^4$ – 1*$(3-1)^4$
= 81 – 16
= 65

Answer:

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