Question 5210 – Data-Structures
January 25, 2024GATE 2003
January 26, 2024Operating-Systems
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Question 2
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- 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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65
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Let the levels be 1,2,3,4,5,6. (1 is the least difficult, 6 is the most difficult level)
Let the computers be F,M,S( fast, medium, slow).
As per the given constraint, 1 must be given to F and 6 must be given to S.
Now we are left with 2,3,4,5 and F being assigned 1, S being assigned 6 and M being assigned none.
Another constraint is that, every computer must be assigned atleast one.
So compute with assigning one job to M, two jobs to M, three jobs to M and four jobs to M.
Assigning one job to M: we can assign 1 out of 4 jobs in (4C1 ways) and remaining 3 jobs to F,S in 2*2*2 = 8 ways. (each job has two options, either F or S),
Assigning two jobs to M: we can select two jobs from 4 in 4C2 ways and remaining 2 can be distributed to F and S in 2*2 ways ( each job has two options either F or S)
Assigning three jobs to M: we can select 3 out of 4 in 4C3 ways remaining can be distributed to F,M in 2 ways.
Assigning 4 jobs to M: it can be done in only one way.
Total : 4c1*8 + 4C2* 4 + 4C3*2 + 1
= 32+24+8+1
=65
Let the levels be 1,2,3,4,5,6. (1 is the least difficult, 6 is the most difficult level)
Let the computers be F,M,S( fast, medium, slow).
As per the given constraint, 1 must be given to F and 6 must be given to S.
Now we are left with 2,3,4,5 and F being assigned 1, S being assigned 6 and M being assigned none.
Another constraint is that, every computer must be assigned atleast one.
So compute with assigning one job to M, two jobs to M, three jobs to M and four jobs to M.
Assigning one job to M: we can assign 1 out of 4 jobs in (4C1 ways) and remaining 3 jobs to F,S in 2*2*2 = 8 ways. (each job has two options, either F or S),
Assigning two jobs to M: we can select two jobs from 4 in 4C2 ways and remaining 2 can be distributed to F and S in 2*2 ways ( each job has two options either F or S)
Assigning three jobs to M: we can select 3 out of 4 in 4C3 ways remaining can be distributed to F,M in 2 ways.
Assigning 4 jobs to M: it can be done in only one way.
Total : 4c1*8 + 4C2* 4 + 4C3*2 + 1
= 32+24+8+1
=65