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Homework #5 — Math 173 — Spring ’
Solutions
(1) 8.1; 8) Let A and B be the sets of families owning a dishwasher and a trash compacter, respectively. We are told that |A| = 50%, |B| = 30% and |A ∩ B| = 20%. We wish to find |A ∪ B| = |U | − |A ∪ B| = 100% − |A ∪ B|. Combining with equation (1) from your text, we have
|A ∪ B| = 100% − |A| − |B| + |A ∩ B| = 100% − 50% − 30% + 20% = 40%.
(2) 8.1; 16) This problem is best done with the Venn diagram. Draw a picture! Let T , B and C stand for the sets liking tennis, bridge and chess, respectively. We are told that |T | = 60%, |B| = 65%, |C| = 50% and
|T ∩ B| = |T ∩ C| = |B ∩ C| = 45%.
If |T ∩ B ∩ C| = 20%, then it follows (look at the diagram you’ve been drawing!) that exactly 25% like exactly each pair of activities. So 25% like exactly C & B, 25% like exactly C & T and 20% like all three. But 25% + 25% + 20% > 50% (50% being the percentage that like C at all). This is a contradiction. If you translate the above logic into equations, where x% is the percentage liking all three, then you see that x% + 2(45 − x)% must be at most 50%. This implies that x ≥ 40. You can check by filling in the rest of the diagram that x = 40 actually works.
(3) 8.2; 8) Let Ai be the set of arrangements in which the letters of type i are adjacent. We wish to count | A¯ 1 ∩ · · · ∩ A¯n|. |U | = (2n)!/(2!)n. Suppose we have an arrangement in the intersection of k of the Ai’s. Then we can imagine ordering the corresponding k pairs along with the 2n − 2 k remaining letters. This can be done in (k + (2n − k))!/(2!)^2 n−k^ ways. Of course, the k letter-types to be paired can be chosen in
(n k
ways. Putting it all together, we get: ∑^ n
k=
(−1)k
n k
(2n − k)! (2!)n−k^
Note:
(2n)! = 2n · (2n − 1) · · · 2 · 1 , 2 n! = 2(n · (n − 1) · · · 2 · 1) = 2n · (2n − 2) · · · 4 · 2.
These are not equal. It may seem a small point, but we don’t write myth when we mean math. They’re different words.
(4) 8.2; 24) First we choose a color for the ceiling in five possible ways. Then we paint the walls of the room with the remaining four colors. Number the walls of the room from 1 to 4. Let A 1 , A 2 , A 3 and A 4 denote the paint schemes in which walls 1&2, 2&3, 3&4 and 4&1 are painted the same color, respectively. It is easy to check that |Ai| = 4^3. For S 2 , there are two cases: Two pairs of disjoint walls, each pair painted the same color, or three walls all painted the same color. In both cases, perhaps surprisingly, you find that |Ai ∩ Aj | = 4^2. For S 3 and S 4 , convince yourself that there are 4 ways to paints the walls for each 3- and 4-intersection. Combining as usual, we find
5 ·
1
2
Note: It is also possible to include the ceiling and define eight sets Ai, one corresponding to each possible wall-wall or wall-ceiling border. However, the number of possible ways to paint an element of, say, Ai ∩Aj ∩Ak depends on i, j, k. For example, if i, j and k correspond to the (wall 1)-(wall 2) border and the (wall 1)-ceiling and (wall 2)-ceiling borders, there are 53 possible ways to paint everything. But if i, j and k all correspond to wall-wall borders, there are only 5^2 ways to paint the surfaces. Additionally, you would need to go all the way out to S 8.
(5) 8.2; 28) This is very similar to the problem with the pairs of letters. There are two main differences. The first is that the members of a couple are presumably distinguishable, so we won’t be dividing by all of the 2!’s; rather, once we choose a place for a couple, we need to multiply by 2 to determine where each member of the couple sits. The second difference is that we are at a round table; i.e., the only thing we care about is who sits next to whom, not who is facing the windows. Now, there are n! arrangements of n distinct objects placed in a line. When we place them in a circle, there are only (n − 1)! arrangements because we identify all n cyclic shifts of a given arrangement with each other. Example, the six permutations of {a, b, c} are {abc, cab, bca} and {acb, bac, cba}. If we interpret these as clockwise seatings around a round table, then the first grouping are all the same single seating and the second grouping are all the same single seating. Now we attack in the similar manner to 8.2; 8). We get:
(2n − 1)! −
n 1
2(2n − 1 − 1)! +
n 2
22 (2n − 1 − 2)! + · · · =
∑^ n
k=
(−1)k
n k
2 k(2n − 1 − k)!.
(6) 8.2; 38) This can be done in several ways. One, which I mentioned in class, is to use Theorem
- Another is to consider the
2
pairs of suits: {HC, HD, HS, CD, CS, DS}. Then define A 1 to be the set of all hands void in Hearts and Clubs; A 2 to be the set of all hands void in Hearts and Diamonds, etc. The following cardinalities can easily be computed: |Ai| =
6
|Ai ∩ Aj | = 0 if Ai ∩ Aj = ∅, |Ai ∩ Aj | =
6
if |Ai ∩ Aj | = 1. You might think that we can ignore S 3 , but it’s not so. For instance, with the ordering above, A 1 ∩ A 2 ∩ A 4 corresponds to all hands void in H,C and D. There are certainly hands that qualify. You can check that there are 4 of these triples that only cover three of the suits. Plugging into our formula, we have: (^) ( 52 6
Note: Several people tried setting Ai to be “void in suit i” for i = 1, 2 , 3 , 4. Then using inclusion/exclusion to compute the number of hands not void in any suit. Then adding in the number of hands void in exactly one suit. This is correct in theory, but it is not true that S 1 − S 2 + S 3 computes the number of hands void in exactly one suit. Rather, it is the number void in at least one suit.
(7) 6.1; 4b) (x^3 + x^4 + x^5 + x^6 )^4.
(8) 6.1; 14) a) (x^1 + x^3 + x^5 )^3 (x^2 + x^4 + x^6 )^3.
b)
∏^6
i=
(x^1 + x^2 + x^3 + x^4 + x^5 + x^6 − xi)
