The World History course and United States history course is already part of the 12 courses he must choose from. It is required to have at least one of these classes in his new class schedule. Sorry if my question didn't make sense earlier, this question was from a practice exam I'm just recalling the question from memory. I was short on time so I just solved C(12,5) and I know its wrong because I forgot about the at least. This has been bugging me all day and I'd appreciate it if someone can explain.
2026-03-27 01:53:38.1774576418
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A student must pick 5 classes from 12 courses, if he must have at least one WH class or USH class, how many different choices does he have?
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There are $\binom{12}{5}$ ways to choose $5$ courses out of the twelve under no restrictions.
There are $\binom{11}{5}$ ways to choose $5$ courses out of $11$ if we don't consider taking World History
There are $\binom{11}{5}$ ways to choose $5$ courses out of $11$ if we don't consider taking Biology
There are $\binom{10}{5}$ ways to choose $5$ courses out of $10$ if we don't consider taking both World History and biology.
Total number of possibilities to choose $5$ courses with World History and biology by using the inclusion and exclusion principle would be $$ \binom{12}{5} - \binom{11}{5} - \binom{11}{5} + \binom{10}{5} = 120 $$
That is if he will have to choose both of them with 3 different courses. However, if he will have to choose 4 courses and one of these two subjects, the answer would be $$ \binom{12}{5} - \binom{10}{5} = 540 $$ However, I am not quite sure.
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We will take "At least one USH or WH" to mean that you could take all five courses from this set. This is odd, but it is the best semantic match for the question wording. If it is supposed to mean one or both, this approach wont work.
In this case, the "stars and bars" technique applies. Ken Ribet is a better teacher than I. If you don't have time for a sixteen minute video, here's a summary:
Your question can be reframed by asking how many ways can you toss k balls into n buckets. Or, how many ways can you insert n - 1 separators into a list of k potential courses. The spaces in between separators represent different buckets (potential courses). There are one fewer separators than buckets because extremal areas count as buckets so the formula is:
${n+k-1\choose k}$
Your question is further complicated by the initial condition of at least one course of two potential types. Now we can count the possibilities by separating into ways of choosing courses from the first two types:
1 History: ${1+2-1\choose1} * {4+10-1\choose4}$ +
2 History: ${2+2-1\choose2} * {3+10-1\choose3}$ +
3 History: ${3+2-1\choose3} * {2+10-1\choose2}$ +
4 History: ${4+2-1\choose4} * {1+10-1\choose1}$ +
5 History: ${5+2-1\choose5}$
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Edit: this answer is solved for a different problem. Specifically I was under the presumption all classes were mandatory rather than only one class being mandatory. I will leave it be however.
Short answer: hold 2 classes fixed for C(10,3)
Tall answer: The answer C(12,5)-C(10,5) is wrong. Counterexample: 2 classes to choose among 4 with 1 biology and 1 world history has one solution. However C(4,2)-C(2,2)=5.
This answer does not remove the combinations C(2,1)C(2,1) which correspond to only one class being chosen.
Also wrong is the Star's and Bar's approach as it presumes there are infinite biology and world history classes. The wording "at least" is very suggestive, however the problem cannot be solved without knowing the actual number of duplicate classes as is seen from observing a simpler case (a technique you should pick up).
The general formula for choosing k classes among n, with m classes being mandatory and not having duplicates is C(n-k,m-k) because you are holding m classes fixed. However following the other line of reasoning of counting up all combinations, subtracting those lacking 1 mandatory class, those lacking 2 mandatory classes, etc...
$$C(n,k)-C(m,m-1)C(n-m,k-m+1)-C(m,m-2)C(n-m,k-m+2)-\cdots-C(m,0)C(n-m,k) = C(n-m,k-m)$$ Because $$\sum_{i = 0}^m C(m,i)C(n-m,k-i)= C(n,k)$$ and is a well known result in combinatorics
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Well there can be another way. We can choose one from World History and Biology and 4 from 10 other classes so the number of ways $C(2,1)C(10,4)$, next we can choose two from World History and Biology and 3 from 10 other classes so the number of ways $C(2,2)C(10,3)$. So total number of ways is $C(2,1)C(10,4) + C(2,2)C(10,3)$.
Note that this method and Henry's method give same result and his method is better in this situation.
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I am supposing that the classes are
$C_1=\text{USH}, C_2=\text{WH}, C_3, C_4, \dots, C_{12}$
]The first class must be either $C_1$ or $C_2$. Having made that choice, he must choose four more classes from the remaining $11$ classes.
There are two ways to choose the first class.
That will leave 11 choices for the next four classes.
Total number of choices is $\binom{2}{1} \times \binom{11}{4}$
Added later as an alternative approach