Math 2001 Homework #10 Solutions
1. Section 4.1: 6ab. For each map below, determine the number of southerly paths from point A to point B.
Solution: We just have to use the same process as we did in building Pascal’s triangle: mark a 1 to count the paths coming directly out of A, then below that at each vertex add up all the paths coming into that vertex. For part (a) we obtain 8 paths, and for part (b) we obtain 10 paths as shown.
A
B
1 1
1 1
2
3 3
8
A
B
1 1
1 1
2
3 3
1 1
4 4
10
2. Section 4.2: 4. For each map below, determine the number of southerly paths fromA to C that pass through B.
Solution: The strategy here is to first figure out how many paths there are from A toB, then figure out how many paths there are fromB toC. We then multiply these possibilities (since once I get to B, it doesn’t matter how I got there when I start heading out to C).
In part (a) we get 10 paths from AtoB using the standard Pascal’s triangle (since it’s a regular grid) which are colored green. Then we start over in red and get another 10 paths from B toC again using Pascal’s triangle. So in total there are 100 paths from A to C that pass through B.
In part (b) we get 14 paths from A to B colored in green. Then starting over in red we get 4 paths fromB toC. So in total there are 56 southerly paths from AtoC that pass through B.
B
C
1 1
1
1 1 2
3 3
4 6
10
1 1
2
1 1
1 3 3
6 4
10
A
B
C
1 1 1
1
1
1 1 2
2 3
3 4
5 6
9
14
1 1
1 1
1 2
2 4
A
3. Section 4.2: 6. A social studies test consists of ten multiple-choice questions, each with five possible choices; ten True/False questions; and ten “matching” questions in which the ten correct answers are given and are to be matched with the appropriate question. (a) How many ways can each section of the test be randomly completed if, in the
matching section, each answer is used exactly once?
Solution: For the multiple-choice portion, there are ten questions and five choices for each question, so there are 510 ways to answer it. For the true/false question, there are ten questions and two choices for each question, so we get 210 ways to complete this. Finally for the matching question, since each answer must be used only once, we have 10! ways of permuting the answers and pairing with the questions.
(b) How many ways may the entire test be randomly completed?
Solution: Just multiply the number of ways to complete each portion: that’s 510·210·10! = 1010·10!.
4. Section 4.2: 8.
(a) How many different license plates are possible if each is to contain three letters followed by three digits?
Solution: There are 26 choices for each of the first three slots, then 10 choices for each of the next three slots, so a total of 263·103 plates.
(b) How many different license plates are possible if each is to contain three letters and three digits in any order?
Solution: To build such a license plate, we would have to first decide which three of the six slots will be letter slots (and all the rest will be number slots). This is the same as the number of subsets of size 3 from the six element set{1,2,3,4,5,6}
Once we’ve decided which slots have letters and which have numbers, the answer is the same as before: 263·103.
So in total we have
6 3
·263·103 possible license plates of this form.
5. Section 4.3: 1egj. Calculate: (e)
50 7
= 50·49·48·47·46·45·44 7·6·5·4·3·2·1 =
50·7·1·47·46·3·44
1 = 99,884,400.
(Here I took a 7 out of 49 to get 7, the 6· 4·2 out of 48 to get 1, and the 3·5 out of 45 to get 3.)
(g)
32 12
= 32·31·30·29·28·27·26·25·24·23·22·21 12·11·10·9·8·7·6·5·4·3·2·1 = 31·29·28·3·26·5·23
= 225,792,840.
Here I factored 832·4 = 1, 1030·3 = 1, 279 = 3, 255 = 5, 1224··721·6 = 1, and 1122·2 = 1. (j)
81 78
= 81·80·79
3·2·1 = 27·40·79 = 85,320. 6. Section 4.3: 3ace.
(a) Find the coefficient of x13y7 in the expansion of (x+y)20. Solution: From the binomial theorem, it’s
20 7
= 20·19·18·17·16·15·14
7·6·5·4·3·2·1 = 19·17·16·15 = 77,520. (b) Find the coefficient of x7 in the expansion of (x−3)12.
Solution: The expansion will have a term like 127x7(−3)5, so the coefficient of x7 is
(−3)5
12 7
=−243· 12·11·10·9·8
5·4·3·2·1 =−243·11·9·8 =−192,456. (c) Find the coefficient of x6 in the expansion of (x+ 2)11.
Solution: As before, the expansion has a term like 116x625, so the coefficient of x6 is
25·
11 6
= 32·11·10·9·8·7
7. Suppose you remove all prime numbers from a standard deck of cards because they’re unlucky, as well as removing the jacks because what’s a jack? So the only faces left are {A,4,6,8,9,10, Q, K}, with all four suits of each.
(a) How many five-card poker hands are there with such a deck? Solution: There are 32 cards and we need to choose 5, so
32 5
= 32·31·30·29·28
5·4·3·2·1 = 8·31·29·28 = 201,376.
(b) How many full house hands are there? (Three of one face, two of another face, such as Aces of clubs, diamonds, and hearts, and 9s of spades and diamonds.) Solution: First we choose the triplet: 81. Then we choose the suits for the triplet: 43. Next we choose the doublet: 71. Then we choose the suits for the doublet: 42. The total number is
8·4·7·6 = 1344.
(c) How many straight flush hands are there? (Five cards in a row, all of the same suit, such as Ace, 4, 6, 8, 9 of hearts. (Note that{A,4,6,8,9}and{9,10, Q, K, A} both count since Ace can be considered either the highest or the lowest card.) Solution: First we choose the starting card (which can be {A,4,6,8,9}), then we choose the suit. In total there are 5·4 = 20 such hands.
(d) How many flush hands are there? (Five cards all of the same suit, but not in a row.)
Solution: We need to choose the suit: 41. Then we choose any five cards in that suit; there are
8 5
= 8·7·6 3·2·1 = 56
ways to do this. Thus in total there are 4·56 = 224 flushes. However 20 of them are straight flushes, so only 204 of them are genuinely flushes.
8. Section 4.5: 3. Refer to the maps of Exercise 4 of Section 4.2. For each map, what is the probability that a randomly selected southerly path from point A to point C passes through point B?
Solution: We have already done half the work here: we counted all the paths from A toC that pass through B. To find the probability, we just need to countall the paths from A toC.
In part (a), we fill in the usual terms of Pascal’s triangle to find 252 paths from A to C. An alternate way of doing it without filling in the triangle is to just use the fact that point C is ten rows down and right in the middle, so it’s
10
Either way, the probability in part (a) of a random path going through point B is 100
252 = 39.7%.
In part (b) we do the nonstandard Pascal’s triangle to get from point A to point C. Adding up at each vertex the numbers from all attached vertices above it, we eventually obtain 151 paths from A to C. Since we already calculated that there are 56 paths that pass throughB, the probability of a random path going through B on the way to C is
56
151 = 37.1%.
C
1 1
1
1 1 2
3 3
4 6
10
A
1
4
10
1 1
1 5 5 1
6 15 20 15 6
21 35 35 21
56 70 56
126 126
252 C
1 1 1
1
1
1 1 2
2 3
3 4
5 6
9
14 A
1
1 2 5 9
14 6
1
6
20 20
23 37 37
57 57
94 151
