# Permutation with restriction [Pending...]

### My question

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

3. Permutations (Ordered Arrangements)

### 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

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

Continues below

## Re: Permutation with restriction

Ioannis

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:

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

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

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.