# Integer Partition of Class Hours, with Restriction

A student has to take $$8$$ hours of classes a week. They want to have fewer hours on Friday than on Thursday. In how many ways can they do this? Assume that classes are only held Monday-Friday and that only whole number hours can be taken on each day.

It's obvious that this is an integer partition problem. We desire an integer partition of $$8$$ into exactly $$5$$ parts (one part representing each day). The number of such partitions is represented by $$p_k(n)$$.

The progress I have made so far is in trying to determine the partitions of $$8$$ into $$5$$ parts, but I am unsure how to take into account the restriction that Thursday must have more hours than Friday.

• I think it has to do with compositions rather than partitions. Taking $1$ hour on Monday and $2$ on Tuesday is different from $2$ hours on Monday and $1$ on Tuesday, I would say. Feb 13, 2020 at 3:17
• @saulspatz Ahh, I see your point. That makes more sense, since those arrangements would constitute different class schedules.
– user746500
Feb 13, 2020 at 3:20

I split it into $$4$$ cases:

1) $$0$$ hours in Friday

2) $$1$$ hour in Friday

3) $$2$$ hours in Friday

4) $$3$$ hours in Friday

(Note that the student can't take more than $$3$$ hours in Friday, or else there will be no way that Thursday$$>$$Friday)

Case 1: $$0$$ hours in Friday

This will just be splitting $$8$$ hours into $$4$$ days, so that will be $$\binom{8+4-1}{4-1}=\binom{11}{3}=165$$, but Thursday can't have $$0$$ hours, so we got to subtract off $$\binom{8+3-1}{3-1}=\binom{10}{2}=45$$. That means that the answer for Case 1 is $$165-45=120$$

Case 2: $$1$$ hour in Friday

As before, we start be splitting off $$7$$ hours($$1$$ of the hours is taken up by Friday) into $$4$$ days: $$\binom{7+4-1}{4-1}=\binom{10}{3}=120$$, but Thursday can't have $$0$$ or $$1$$ hours, so we have to subtract off $$\binom{7+3-1}{3-1}=\binom{9}{2}=36$$ for Thursday having $$0$$ hours and $$\binom{6+3-1}{3-1}=\binom{8}{2}=28$$ for Thursday having $$1$$ hour. That will be $$120-36-28=56$$ for Case 2.

Case 3: $$2$$ hours in Friday

We start by splitting off $$6$$ houts into $$4$$ days, so that will be $$\binom{6+4-1}{4-1}=\binom{9}{3}=84$$. Thursday can't be $$0, 1, \text{ or }2$$ hours, so we will have to subtract

$$\binom{6+3-1}{3-1}=\binom{8}{2}=28$$, for Thursday $$=0$$ hours

$$\binom{5+3-1}{3-1}=\binom{7}{2}=21$$, for Thursday $$=1$$ hours

$$\binom{4+3-1}{3-1}=\binom{6}{2}=15$$, for Thursday $$=2$$ hours

That means that Case 3 will be $$84-28-21-15=20$$

Case 4: $$3$$ hours in Friday

Thursday can only be $$4$$ or $$5$$ hours, so it will be $$1$$ hour into $$3$$ days which will result in $$3$$ ways, and only $$1$$ way(Monday to Wednesday$$=0$$, $$5$$ hours for Thursday and $$3$$ hours for Friday) for Thursday $$=5$$ hours. That will be $$3+1=4$$ for Case 4

In total that will be: $$120+56+20+4=200$$ ways.

(Not sure if this is correct; I am only in the 8th grade so I'm quite new to these kind of stuff)

(I used Stars and Bars btw if you didn't know)

Let me get you started. The student can take no more than $$3$$ hours on Friday. Suppose he takes $$0$$ hours on Friday. Then he must take at least $$1$$ hour on Thursday. That leaves $$7$$ hours to be distributed from Monday through Thursday, and this is a stars and bars problem with $$\binom{10}{3}$$ solutions.

Do you see how to continue?

Note that we can either have more hours on Friday than Thursday, less hours, or the same number of hours. The number of distributions of hours that give us more hours on Friday than Thursday is the same as the number of distributions giving us less hours on Friday (i.e. we have a bijection between the sets by switching the hours on Friday/Thursday). So we will count the number in either of these sets by finding the total number of distributions, subtracting the number of distributions where Thursday and Friday have the same number of hours, and then divide by $$2$$.

So we have $$x$$ hours on Friday and Thursday. So we have $$8-2x$$ hours to distribute on the other $$3$$ days, so in this case we have $${10-2x \choose 2}$$ choices by a stars and bars argument. Thus we have $$\displaystyle\sum_{x=0}^{x=4} {10-2x \choose 2}$$ for distrubtions with equal hours on Friday and Thursday, and the total number of different distributions is $${12 \choose 4}$$, so we have our answer is $$\frac{1}{2}\left({12 \choose 4} - \sum_{x=0}^{x=4} {10-2x \choose 2}\right) = 200$$