Raoul's Workout Schedule

There are 7 days in a week. Raoul wants to work out 3 days a week.

$7-3 = 4$

So he gets 4 days of rest each week. He wants each visit to be followed by at least one day of rest, but the division $\frac {4}{3}$ has a remainder of 1. So there is one lucky visit that is followed by 2 days of rest. Creating the schedule comes down to choosing which day of the week the lucky visit is. So, there are : ${7 \choose 1}$ = 7 different ways Raoul can schedule his visits to the gym.

Keeping in mind the constraints given, lets start working out the solution from monday. The possible combinations are MON, WED , FRI.........(1) ( where it represents the days on which Raoul goes to gym). The other ones are MON,WED,SAT ..................................(2)& MON,THU,SAT .....................................(3)

For Tuesday, TUE,THU,SAT ..................................(4) TUE,THU,SUN ....................................(5) TUE,FRI,SUN ......................................(6)

For Wednesday, WED,FRI,SUN ......................................(7) WED,FRI,MON ......................................(8) WED,SAT,MON .....................................(9)

But we see that (8) and (9) are same as (1) and (2) respectively. So they are neglected. Similarly, there are repititions for other days of the week as well. Therefore we get the final answer as 7.

If we need to schedule 3 visits in 7 days so that there is at least 1 day between each visit, there should be 2 one-day breaks and 1 two-day break (three-day break is impossible, which can be easily checked).

We continue by considering 3 cases of possible ordering of these breaks in a week starting from the first day of visit:

I. One-day, one-day, two-day -- possible schedules are (by number of day of visit) 1-3-5; 2-4-6; 3-5-7

II. One-day, two-day, one-day -- possible schedules are 1-3-6; 2-4-7

III. Two-day, one-day, one-day -- possible schedules are 1-4-6; 2-5-7

In total there are 7 different ways to schedule gym visits under given conditions.

Since we know that number of ways to select r objects from n distinct objects is nCr, which is = n!/((n-r)!r!), ways to select 3 days from a week of 7 days is 7C3=7 * 6 * 5/(1 * 2 * 3) = 35 Invalid selections are 3 consecutive days and 2 consecutive days and 1 different day. Number of selections of 3 consecutive days are 7. Number of selections of 2 consecutive days are 7, and no. of different non consecutive days are 7*3=21 Therefore number of valid selections is 35-(7+21)=7

(Sat , Sun , Mon , Tue , Wed , Thu , Fri , Sat , Sun , Mon , Tue , Wed , Thu , Fri , ..……….) Let , a = 1st day , b = 2nd day , c = 3rd day … now, if a = Sat
Then, b ≠ (Sat , Sun , Wed , Thu , Fri) . And c ≠ (Sat , Sun , Mon , Tue , Fri) [Note : when a = Sat , if b = (Sat) then a = b . if b = (Sun) then there will be no gap between a and b . If b = (Fri / Thu) then there is no room for c . And if b = (Wed) then c must be Fri . but if c = (Fri) then there is no gap between a and c . So, b ≠ (Sat , Sun , Wed , Thu , Fri) . For same kind of reasons c ≠ (Sat , Sun , Mon , Tue , Fri) ] so, if b = (Mon) then, c =(Wed / Thu). and if b =(Tue) then, c =(Thu). so the possible answers are 1.( Sat , Mon , Wed) , 2.(Sat , Mon , Thu) , 3.(Sat , Tue , Thu ).

Now, if a = Sun . then for similar kind of reasons
b ≠ (Sun , Mon , Thu , Fri , Sat) and, c ≠ (Sun , Mon , Tue , Wed , Sat ) by the same process , the possible answers are 4.(Sun , Tue , Thu) , 5.(Sun , Tue , Fri) , 6.(Sun , Wed , Fri) again , by the same process we can see when c = Mon the possible answer are 7.(Mon , Wed , Fri) , 8.(Mon ,Wen , Sat) , 9.(Mon , Thu , Sat) . Now , we can see that answer No. 8 and 1 are same as well as No. 9 and 2.( Just there starting day is different ) . Now, if we go with the same process we will see that the first (7) answers will came again and again as a circle . So there is only (7) distinct answers .

Here we assume that the start of the week is on Monday and the end of the week is Sunday. If his first visit of the week is on Monday, there will be 3 ways to schedule the other 2 visits, namely (Wednesday,Friday) , (Wednesday,Saturday) , (Thursday,Saturday) , his third visit can't be on Sunday. Similarly, if his first visit is on Tuesday, there will be 3 ways to schedule the other 2, namely (Thursday,Saturday) , (Thursday,Sunday) , (Friday,Sunday). If his first visit is on Wednesday, the other 2 visits must be on Friday and Sunday. Finally, if his first visit is Thursday or later, he can't have at least 1 day in between his gym visits. Therefore we have 3+3+1 = 7 ways to schedule his gym visits.

A week have Sun, Mon, Tue, Wed, Thu, Fri, Sat. Begin from Sun: we have Sun-Tue-Thu, Sun-Tue-Fri, Sun-Wed-Fri (there aren't Sun-Tue-Sat, Sun-Wed-Sat and Sun-Thu-Sat since the next day of Sat is Sun) Begin from Mon: we have Mon-Wed-Fri, Mon-Wed-Sat, Mon-Thu-Sat Begin from Tue: we have Tue-Thu-Sat Those above make 7 ways. We can't begin from Wed since Wed-Fri-Sat or Wed-Thu-Sat don't meet the requirement having at least 1 day in between each visits

First, let the days of the week be replaced with numbers (i.e Monday = 1, Tuesday = 2, ... , Sunday = 7) and hence let the days of the week be Monday to Sunday in that order.

Since there must be at least one day between each visit, we can see that it is the same as adding 2 to the number corresponding to the day (i.e 1 + 2 = 3, Wednesday).

Aside from that we can also see that if the first workout of the week is on a Monday, then the last can't be a Sunday or else there won't be a day between the two work outs.

Also, since the minimum total sum is 2+2=4, and the highest day we have is Sunday or 7, the latest day the first day can be is therefore Wednesday.

If the first day is a Monday, we can add a total of 5, and a minimum of 4, so the different combinations are 2+2, 2+3, or 3+2 which gives three schedules.

If the first day is Tuesday then the largest total sum is 7-2=5, and the minimum is again 4, so we have 2+2, 2+3, or 3+2 which gives another three schedules.

Finally, if Wednesday is the first day, we can add a maximum of 7-3=4, and since the minimum is also 4, there is only one schedule starting on a Wednesday (i.e 2+2).

So we have three schedules starting on a Monday, three schedules starting on a Tuesday, and one starting on Wednesday which gives a total of 3+3+1=7 different schedules.

Solution 1: Since there are 7 days in a week, and Raoul goes to the gym 3 of these days, there must be 4 days in each week where he does not go to the gym. Since there are more days when he does not go to the gym, we must be able to find a pair of consecutive days where Raoul does not go to the gym. Based on the number of days Raoul is at the gym, there can be at most one pair of consecutive days where Raoul does not go to the gym. Thus, Raoul's gym schedule is a cyclic permutation of $1,0,1,0,1,0,0$ where $1$ represents the days he goes to the gym and $0$ represents the days he does not. There are 7 cyclic permutations of this schedule, so there are 7 ways for Raoul to select his schedule.

Solution 2: We count the number of viable schedules using PIE. There are $\binom{7}{3} = 35$ ways to choose 3 days of the week for Raoul to go to the gym. By considering some pair of gym days to be consecutive, we see there are $\binom{6}{1}\binom{5}{1} = 30$ ways to choose 3 days of the week with some pair consecutive. By considering all three gym days to be consecutive, we see there are $\binom{5}{1} = 5$ ways to choose 3 days of the week with all days consecutive. This gives a total of $35 - 30 + 5 = 10$ ways to make the schedule. However, a schedule generated by this may have both the first and last day of the weeks as gym days. Since Raoul wants to visit the gym the same 3 days each week, this cannot occur. Any schedule where this occurs has day 1 and 7 as gym days, and one of days $3,4,5$ as the final gym day, so there are 3 of them in total. Thus, there are only 7 ways Raoul can schedule his workouts.