Winston's Semester

There are 10 total classes offered, so there are ${10 \choose 4} = 210$ potential choices. However, not all of them include at least 1 science class and at least 1 arts class. There are ${6 \choose 4} = 15$ choices with no science classes and ${7 \choose 4} = 35$ choices with no arts classes. There is no way to take 4 classes with no science or arts, so these two groups don't overlap. Excluding these two groups, all the remaining choices have at least one art and at least one science. Therefore, there are $210 - 15 - 35 = 160$ choices.

You might like to try an Inclusion/Exclusion type counting method, but since the set is small enough I just listed the number of ways he can take x science and y art classes

(x,y): # of ways

(1,1):4*3*3

(1,2):4*3*3

(1,3):4

(2,1):6*3*3

(2,2):6*3

(3,1):4*3

for a total of 160

He has 6 ways to choose his classes: _1 science class, 1 art class, 2 other classes: there are 4 \times 3 \times 3 = 36 (choices). _ 2 science classes, 1 art class, 1 other class: there are 6 \times 3 \times 3 = 54 (choices). _2 science classes, 2 art classes: there are 6 \times 3 = 18 (choices). _1 science classes, 2 art classes, 1 other class: there are 4 \times 3 \times 3 = 36 (choices). _1 science class, 3 art classes: there are 4 \times 1 = 4 (choices). _3 science classes, 1 art class: there are 4 \times 3 = 12 (choices). Therefore, the total choices for classes Winston has are: 36 + 54 +18 + 36 + 4 + 12 = 160 (choices).

Let the number of classes Winston chooses in Science, Art, and Other be (x,y,z). Then the number of choices for a certain ordered triplet is equal to f(x,y,z) = 4Cx + 3Cy+ 3Cz, where C denotes combinations. The six possible ordered pairs can easily be listed: (1,1,2); (1,2,1); (1,3,0); (2,1,1); (3,1,0); (2,2,0). These respectively give the values 36, 4, 54, 12, and 18, which sum to 160, WWWWW.

If there are no restrictions of the choice of classes (so we don't have to choose at least one science class and one arts class), then there are ${10 \choose 4}$ choices for 4 classes.

If we do not choose any science classes, then there are ${6 \choose 4}$ choices for 4 classes.

since there are only 3 other classes, if we choose 4 classes out of the 6, the resulting combination will always have an art class.

If we do not choose any art classes, then there are ${7 \choose 4}$ choices for 4 classes.

since there are only 3 other classes, if we choose 4 classes out of the 7, the resulting combination will always have a science class.

So there are ${6 \choose 4}$ + ${7 \choose 4}$ combinations that do not have an art class or a science class.

So there are ${10 \choose 4}$ - ${7 \choose 4}$ - ${6 \choose 4}$ =160 combinations that have an art class and a science class.

The answer is 160.

We name the science classes $S_1$ through $S_4$ , the art classes $A_1$ through $A_3$ , and the other classes $O_1$ through $O_3$ . We have the set of classes $C=\{S_1,S_2,S_3,S_4,A_1,A_2,A_3,O_1,O_2,O_3\}$ . From these, we are trying to select 4 classes, giving a gross total of $\binom{10}{4}=210$ choices. However, there are several cases that we must omit as they do not satisfy the given requirements.

Case 1: One science class and all three of the "other" classes are selected.

By the fundamental counting principle, this produces $\binom{4}{1}\binom{3}{3}=4$ outcomes that we must omit.

Case 2: One art class and all three of the "other" classes are selected.

By the fundamental counting principle, this produces $\binom{3}{1}\binom{3}{3}=3$ outcomes that we must omit.

Case 3: One "other" class and three science classes are selected.

By the fundamental counting principle, this produces $\binom{3}{1}\binom{4}{3}=12$ outcomes that we must omit.

Case 4: One "other" class and all three art classes are selected.

By the fundamental counting principle, this produces $\binom{3}{1}\binom{3}{3}=3$ outcomes that we must omit.

Case 5: Two "other" classes and two art classes are selected.

By the fundamental counting principle, this produces $\binom{3}{2}\binom{3}{2}=9$ outcomes that we must omit.

Case 6: Two "other" classes and two science classes are selected.

By the fundamental counting principle, this produces $\binom{3}{2}\binom{4}{2}=18$ outcomes that we must omit.

Case 7: All four science classes are selected.

This produces $\binom{4}{4}=1$ outcome that we must omit.

Totalling up the rejected cases, we get 50 unwanted cases. Subtracting this from the total we obtain $\boxed{160}$ possible ways for Winston to choose his classes. How can we be sure that we have counted all possible rejected cases? If you check any of the cases above, replacing a class type for one class either produces another listed rejected case or a supported case. Thus we can be sure we have counted all cases.

Let S, A, and O denote a science class, art class, and other class respectively. Then there are 6 possible schedules for Winston to choose that satisfy the minimum science and art requirements. They are: 1. S A S S (i.e. 3 science classes and 1 art class) There are {4 \choose 3} = 4 ways to choose the science classes and 3 art classes to choose from, thus there are 4 \cdot 3 = 12 schedules with 3 science classes and 1 art class. I will go into less detail from here on; the arguments for the remaining schedules are the same. 2. S A S A {4 \choose 2} \cdot {3 \choose 2} = 18 3. S A S O {4 \choose 2} \cdot {3 \choose 1} \cdot {3 \choose 1} = 54 4. S A O O {4 \choose 1} \cdot {3 \choose 1} \cdot {3 \choose 2} = 36 5. S A A O {4 \choose 1} \cdot {3 \choose 2} \cdot {3 \choose 1} = 36 6. S A A A {4 \choose 1} \cdot {3 \choose 3} = 12

Adding up all these possibilities gives the answer, 160 different schedules.

Let us first count how many ways Winston can choose his classes without restriction. This is given by 10 choose 4. Which is 210.

Let us now compute for the complement of what we are asked to find. The complement of taking at least 1 science class and at least 1 arts class is taking no science class or taking no arts class.

If there will be no science class, we will only have 6 options to choose 4 classes from. This is given by 6 choose 4. Which is 15.

If there will be no arts class, we will only have 7 options to choose 4 from. This is given by 7 choose 4. Which is 35.

Adding these two cases together will give us:

15 + 35 = 50

Now we have to subtract this sum from the number of ways to choose classes without restriction so we can get the number of ways to take at least 1 science class and at least 1 arts class.

210 - 50 = 160