I have the following problem. Three irrigation events should implemented in a 19days period. Between 2 irrigation events there should be at least a 5 days interval (i.e. if the first irrigation is implemented in day 1 the second should take place at day 6 or latter). No irrigation should be scheduled later than day 19. How can I calculate the maximum number of irrigation schedules that meet the above mentioned restrictions? I have solved the problem by hand and I know that they are 35, but is there any mathematical formula that solves the problem?

I thank is advance anyone who would spend time to read the problem and give an answer

Initially I created the first irrigation schedule like:
First Irrigation at day 1
Second Irrigation at day 1+5=6
and third Irrigation at day 6+5=11
Based on problem restrictions the first irrigation could happen at days 1-2-3-4-5, the second at days 6-7-8-9-10 and the third at days 11-12-13-14-15. So we have three events with 5 elements each. If there were no restrictions the answer will be 5x5x5=125 different schedules. However, I could not figure out how to approach the problem in order to solve it when I have to consider the restrictions

X

Hello everybody.
I have the following problem. Three irrigation events should implemented in a 19days period. Between 2 irrigation events there should be at least a 5 days interval (i.e. if the first irrigation is implemented in day 1 the second should take place at day 6 or latter). No irrigation should be scheduled later than day 19. How can I calculate the maximum number of irrigation schedules that meet the above mentioned restrictions? I have solved the problem by hand and I know that they are 35, but is there any mathematical formula that solves the problem?
I thank is advance anyone who would spend time to read the problem and give an answer
Ioannis

Relevant page
<a href="https://www.intmath.com/counting-probability/3-permutations.php">3. Permutations (Ordered Arrangements)</a>
What I've done so far
Initially I created the first irrigation schedule like:
First Irrigation at day 1
Second Irrigation at day 1+5=6
and third Irrigation at day 6+5=11
Based on problem restrictions the first irrigation could happen at days 1-2-3-4-5, the second at days 6-7-8-9-10 and the third at days 11-12-13-14-15. So we have three events with 5 elements each. If there were no restrictions the answer will be 5x5x5=125 different schedules. However, I could not figure out how to approach the problem in order to solve it when I have to consider the restrictions

So altogether for event 1 being on day 1, we have 45 event 2 and event 3 possibilities.

Then we need to consider event 1 being on day 2, then day 3 and so on.

After we get this "by hand" sorted out, we'll try to come up with a formula.

X

Ioannis
Sorry about the long delay. Maybe you've already figured this out...?
If not, let's work on your given "by hand" solution (of 35), to give you some hints about how to approach it.
I believe it's rather larger than 35.
For example, if the first irrigation event is day 1, the second is day 6, then there are 9 possible events for the 3rd (days 11 through to 19).
If the second event is day 7, there are 8 possible days for event 3 (days 12 to 19)
If we consider days 8 to 14 for event 2, then the possible days for event 3 are:
<pre>Evt 1 Evt 2 Evt 3 poss
1 8 7
1 9 6
1 10 5
1 11 4
1 12 3
1 13 2
1 14 1</pre>
So altogether for event 1 being on day 1, we have 45 event 2 and event 3 possibilities.
Then we need to consider event 1 being on day 2, then day 3 and so on.
After we get this "by hand" sorted out, we'll try to come up with a formula.

I think its too late. However if you are interested you can have a look on answer .

According to Q , it is necessary to have first irrigation and later irrigation can be done in 5 days or later.

x,y,z representing the date of irrigation,the x=1,y,z greater than or equal to 5

15=>x+y+z>=19

Let a dummy variable u where u can vary between 0 and 4(As x,y,z can vary between 15 and 19)

x+y+z+u=19
Since x=1,y>=5,z>=5
Then y+z+u=19-1-5-5=8 (Now y,z,u can vary between 0 and 5)
Here n=0,r=3
Using Combination of selection of 0 or more = (n+r-1)C(r-1)=10C2=45

X

I think its too late. However if you are interested you can have a look on answer .
According to Q , it is necessary to have first irrigation and later irrigation can be done in 5 days or later.
x,y,z representing the date of irrigation,the x=1,y,z greater than or equal to 5
15=>x+y+z>=19
Let a dummy variable u where u can vary between 0 and 4(As x,y,z can vary between 15 and 19)
x+y+z+u=19
Since x=1,y>=5,z>=5
Then y+z+u=19-1-5-5=8 (Now y,z,u can vary between 0 and 5)
Here n=0,r=3
Using Combination of selection of 0 or more = (n+r-1)C(r-1)=10C2=45

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