Counting Methods with Combinatorics Problems (2) I have a couple of problems with my attempts at them. I was wondering if I could get some verification/help if possible!


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*A college sent a report saying that 119 students took Calc. I in a Fall Semester. The report says that during the next term, 96 of those students were in Calc. II, 53 of them took Discrete Mathematics, and 39 of them took Physics II. The report also says that 38 of the students took both Calc. II and Discrete Mathematics, 31 of the students took both Discrete Mathematics and Physics II, 32 of the students took both Calc. II and Physics II, and 22 of the students took all three courses. We examine the report and sense an error is present. Why?


119 students total from the Fall semester. 
96-Calc II 
53-Discrete Mathematics 
39-Physics II 
38-Calc II and Discrete 
31-Discrete and Physics II
32-Calc II and Physics  II 
22-Calc II, Discrete, and Physics II
My guess is that if 22 students took all three courses, that would leave a remainder of 74 that are taking Calc II, 31 taking Discrete Math, and 17 taking Physics II which adds up to 122 students. 122>119. Is that the error? Or am I approaching this wrong?

2. We are going to count banana splits: These are ice cream treats that have 3 scoops of ice cream (two or three of the scoops could be the same flavor), three toppings (two or three could be the same flavor), whipped cream (always), a choice of nuts or no nuts, and a choice of a cherry or no cherry, all placed atop two banana halves. If there are 18 different flavors of ice cream and 5 choices of toppings, how many different banana split orders are possible? Note that people do care which toppings end up on which scoops, so the positions of the scoops should be labeled. 

So, we have to have 3 scoops of ice cream that have 18 different flavors and they can be repeated  3 toppings that have 5 different kinds and they can be repeated  always whipped cream  nuts or no nuts  and a cherry or no cherry. 

I'm going to take a wild shot at this, and I'm probably overthinking it. Order matters and repetition is allowed for the toppings on the scoops. So, we have 18 flavors for our 3 scoops of ice cream. We would use $k^n= 18^3= 5,832$ different scoop & flavor combinations. Same approach with the toppings: $5^3=125$ different combinations. 
I attempted to go further with this, but yielded an incredibly high number...so, I'm stopping here. I think I'm doing something wrong. Any suggestions are appreciated very much! 
 A: 
A college sent a report saying that $119$ students took Calc. I in a Fall Semester. The report says that during the next term, $96$ of those students were in Calc. II, $53$ of them took Discrete Mathematics, and $39$ of them took Physics II. The report also says that $38$ of the students took both Calc. II and Discrete Mathematics, $31$ of the students took both Discrete Mathematics and Physics II, $32$ of the students took both Calc. II and Physics II, and $22$ of the students took all three courses. We examine the report and sense an error is present. Why?

We are told that $22$ students take all three courses, $31$ students take both Physics II and Discrete Mathematics, and $32$ students take both Physics II and Calculus II.  Since $22$ students take all three courses, that means $31 - 22 = 9$ students take Physics II and Discrete Mathematics but not Calculus II and that $32 - 22 = 10$ students take Physics II and Calculus II but not Discrete Mathematics.  If all these numbers are correct, there are at least $22 + 9 + 10 = 41$ students in Physics II.  However, we are also told that a total of $39$ students are enrolled in Physics II, which is a contradiction.   

We are going to count banana splits: These are ice cream treats that have 3 scoops of ice cream (two or three of the scoops could be the same flavor), three toppings (two or three could be the same flavor), whipped cream (always), a choice of nuts or no nuts, and a choice of a cherry or no cherry, all placed atop two banana halves. If there are 18 different flavors of ice cream and 5 choices of toppings, how many different banana split orders are possible? Note that people do care which toppings end up on which scoops, so the positions of the scoops should be labeled. 

What you have done thus far is correct.  Since whipped cream is automatically included, you have to multiply your result by the number of choices entailed in choosing whether or not to order nuts and whether or not to order a cherry on top.
A: For the banana split problem, one approach is to first create a compound item.
In this case, the compound item would would be a scoop with the flavor, toppings, nuts and cherries already pre-selected.
There are $18*5*2*2 = 360$ compound items.
So, the problem simplifies to selecting $3$ times with replacement from $360$ items.
If the order in which you select the compound items matter, there are $360^3$ ways.  
If the order does not matter and all you care about is which $3$ items you end up with, then it gets a bit more complicated and you will have to split it up into cases.
A: First
Indeed an error is present.
Let's call $A$ the set of students who took Calculus I.
Let $C,D,P$ be three of its subsets. (for Calculus, Discrete and Physics)
(I notate the disjoint union $\sqcup$)
Provided information
$$|A|=119, ~~~~~~ |C|=96, ~~~~~~ |D| = 53, ~~~~~~ |P|=39$$
$$|P\cap C| = 32, ~~~~~~ |P\cap D|=31, ~~~~~~ |C\cap D| = 38, ~~~~~~ |C\cap D\cap P| = 22$$
But
$$\matrix{&P\cap D &= &(P\cap D \cap C) & \sqcup & (P\cap D\cap\bar C)\\
\Rightarrow & |P\cap D| & = & |P\cap D \cap C| & + & |P\cap D\cap\bar C|\\
\Leftrightarrow & |P\cap D| & - & |P\cap D \cap C| & = & |P\cap D\cap\bar C|\\
\Rightarrow & 31 & - & 22 & = & |P\cap D\cap\bar C|\\
\Rightarrow & &&|P\cap D\cap\bar C| & = & 9}$$
With a similar reasoning, we find that $~~|P\cap C\cap \bar D|= 10$
Finally,
$$\matrix{&P&=&(P\cap D \cap C)&\sqcup&(P\cap D \cap \bar C)&\sqcup&(P\cap \bar D \cap C)&\sqcup&(P\cap \bar D \cap \bar C)\\
\Rightarrow & |P| & \geqslant&|P\cap D \cap C|&+&|P\cap D \cap \bar C|&+&|P\cap \bar D \cap C|\\
\Rightarrow & 39 & \geqslant&22&+&9&+&10\\
\Leftrightarrow & 39 & \geqslant & 41}$$
Seems like we got ourselves a contradiction.
Solution
$$|P\cap C \cap D| \leftarrow 24$$ 
(I verified every possible relation and there is no contradiction)
Second


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*The choice of scoop flavor is unordered with repetition so the possibilities are $\binom{n+k-1}{k}=\binom{18+3-1}{3}=1140$

*The choice of topping is ordered and with repetition, thous $5^3 = 125$ possibilities

*Nuts and cherries both multiply by 2


Total
We got a total of 570 000 possible banana splits !!
