Solutions to Practice Problems #1 ppg. 12-13 ---------- 3) a) This fits the description of the product rule, since each distinct car is a combination of the four properties. Hence there are a total of 4x12x3x2 = 288 distinct Buicks that can be manufactured. b) If you don't have a choice of the color, as with this question, you can only choose the model, engine size & transimission for 4x3x2 = 24 distinct blue Buicks. 9) a) There are 14 choices for the bakery item. There are 12 choices for a medium beverage (all of these are distinctly listed. Thus, there is a total of 14x12 = 168 total combinations of of one bakery item + medium beverage that Carol can get. b) There are 14 choices for Carol's bakery item. There are 4x3=12 choices for coffee, of any size. There are 6x3=18 choices for tea, of any size. The are 6 choices for Ms. Didio's muffin. Using the product rule, the answer is 14x12x18x6 = 18,144 possible orders. Now, there is a different way to interpret this question. In our first interpretation, we are distinguishing between these two orders: muffin a and black coffee for Carol & muffin b and tea for Ms. Didio muffin b and black coffee for Carol & muffin a and tea for Ms. Didio. We have counted both of these. But, in real life, if Carol was picking up the whole order herself, then these two should NOT both be counted. If you used the second interpretation of the question, it becomes more difficult. Here is that answer: We need to figure out the total number of bakery combinations. We can split these into two groups: combos WITH one pastry, and combos WITHOUT a pastry: Number of combos with a pastry = 8x6 = 48. 8 choices for the pastry and 6 for the muffin. Number of combos with no pastry = ((2+6-1) C 2) = 21. You have to make two choices of pastries out of six, allowing for repetition (from section 1.4). Thus, the total number of food orders is the 48+21 = 69. With this interpretation, the final answer = 18x12x69 = 14,904. 17) We must sum up the total number of 1 letter identifiers, 2 letter identifiers, ... and 8 letter identifiers. Of these, we have counted 36 that we should not have, so we need to subtract 36 from our answer. Use the product rule for each of the independent parts. # of one letter IDs = 26 (1 for each letter) # of two letter IDs = 26x36 (since there are 36 indepedent choices for the second letter.) # of three letter IDs = 26x36x36 (since there are also 36 choices for the third letter.) ... So, continuing this, we find the final answer to be: 26 + 26x36 + 26x36^2 + 26x36^3 + 26x36^4 + 26x36^5 + 26x36^6 + 26x36^7 - 36 = 26x36 + 26x36^2 + 26x36^3 + 26x36^4 + 26x36^5 + 26x36^6 + 26x36^7 - 10 (Note: 36^4 means 36 raised to the power 4, for example.) 23) Imagine that we write down the order of the colors that Deborah breaks her targets. We might write something like RWWGBBWGRRRB. This, completely specifies which targest she hits, in which order. When we see the first R, we KNOW that she broke the lowest red target. In essence, each permutation of the letters RRRRWWWGGBBB corresponds to an unique way to break the targets. Similarly, each way to break the targets uniquely corresponds to a string of Rs, Ws, Gs, and Bs. Thus, there is a one-to-one correspondence between the two counts. The total number of permuations of RRRRWWWGGBBB is 12!/(4!3!2!3!) = 277,200, the answer to the original question. 33) a) 2 choices for each question: 2^10 b) Now, blank counts as an answer for each question. Thus, there are 3 choices for each answer so the total number of responses = 3^10. ppg. 25-27 ---------- 7) a) (20 C 12), since you are choosing 12 members out of 20. b) (10 C 6)x(10 x 6), since you are independently choosing 6 of 10 women and 6 of 10 men. c) You can have 2, 4, 6, 8 or 10 women on the committee. Count the number of ways to do each of these and add them up. Using the logic from part b repeatedly, we get the following answer: (10 C 2)x(10 C 10) + (10 C 4)x(10 C 8) + (10 C 6)x(10 C 6) + (10 C 8)x(10 C 4) + (10 C 10)x(10 C 2) (Note that in each product, the first term stands for the number of ways to pick the women, and the second the men.) d) Same idea as c, but this time add up the number of ways you can have 7, 8, 9 or 10 women on the committee: (10 C 7)x(10 C 5) + (10 C 8)x(10 C 4) + (10 C 9)x(10 C 3) + (10 C 10)x(10 C 2) (Note that in each product, the first term stands for the number of ways to pick the women, and the second the men.) e) Same idea as d, but this time add up the number of ways you can have 8, 9, or 10 men: (10 C 4)x(10 C 8) + (10 C 3)x(10 C 9) + (10 C 2)x(10 C 10) (Note that in each product, the first term stands for the number of ways to pick the women, and the second the men.) 13) This is very similar to the TALLAHASSEE example in the book. Here we have 7 letters: M, I, I, I, I, P, and P to permute. There are 7!/(4!2!) ways to do this. Consider one of these: __ M __ I __ I __ I __ I __ P __ P __ We must place 4 Ss in the 8 __ (blanks) laid out. In particular we must choose EXACTLY 4 of the 8 blanks to place our Ss, since none of them can be consecutive. By separating the blanks with other letters in the word we ensure this. There are (8 C 4) ways to place the Ss for this one permutation. In fact, for each of the different permutations of MIIIIPP, there are (8 C 4) ways to place the Ss. Thus, we can use the product rule to come up with our final answer = 7!/(4!2!)*(8 C 4) = 7350. ppg. 34-35 ---------- 3) We have 4 containers, and 20 coins, plugging directly into the formula on page 30, we have ((20+4-1) C (4-1)) = (23 C 3). 5) a) This is a fairly clever question. The key is to realize the following: You may choose ANY subset of the five distinct coins in the group. Once you have made this choice, (for example, choosing a nickel and a dime), you are FORCED to choose 3 Susan Anthony dollars. Thus, the question really boils down to how many combinations of the first five coins can you have? From before, you can either have the penny or not, two choices. You can either have the nickel, or not, two choices... In total, you have a possible 2x2x2x2x2 = 32 possible combos of the first 5 coins. This is ALSO the total number of ways you can select 5 of the total of 10 coins. b) Using the exact reasoning as part a, the question boils down to how many combos of the first n objects can we have. Using the same reasoning in part a, we find this answer to be 2^n. (2^n simply means 2 raised to the power n.)