Test #2: Possible Solutions

1.
(a) P(20,6)/6!

(b) 10!/(3!3!2!)

(c) 19!/(2!9!5!3!)

(d) C(9,4)

(e) 2^19

2. 4!*P(5,3)
Place the vowels first. There are 4! ways to do this. There are now five places among the vowels into which consonants may be placed so as to keep them from each other. Arrange the 3 consonants among the 5 available spots, done in P(5,3) ways.

3. Here is the expansion.

4. 50
Consider two cases depending on the number of different colors in a flag.

Case I: Flags with three different colors
There are P(5,3) ways to create these flags, but because there is no distinction between the top and bottom of a flag, we have twice as many flags as exist. We divide by 2 to account for flags of, say Red-Green-Blue looking the same as Blue Green-Red. This gives us 30 different flags.

Case II: Flags with two different colors
The only way to have two colors is for the outside stripes to be the same color and the middle color different from the outside. There are P(5,2) ways to create this type of flag. No division by 2 is necessary because of the symmetry of the colors. Red-Green-Red has only been counted once, for example. This yields 20 more flags.

We add the results because Case I and Case II represent two non-overlapping sets: One flag cannot have two different colors and three different colors.

5.
(a) 2^12
For each of the 12 questions there are two choices, YES or NO. We use the multiplication principle because each choice is, we assume, independent from the rest.

(b) C(12,5) = C(12,7)
From among the 12 questions, choose 5 for YES or, equivalently, choose 7 for NO. There is no accounting for the arrangement here, for we are simply gathering 5 (or 7) questions.

(c) 3^12
For each of the 12 questions there are now three choices, YES or NO or ABSTAIN. We use the multiplication principle because each choice is, we assume, independent from the rest.

6. 196 dominoes
There are several ways to solve the problem. One approach is to count the dominoes that can be created for each of three cases, depending on how many different numbers are on the domino, and then add the results.

Case I: Dominoes that show exactly one unique number (7 dominoes)
There is one and only one way to arrange one number three times on a domino. There are seven numbers to choose from, but once a choice is made for one of the spots on a domino, the other two are determined. There are, then, P(7,1) = 7 such dominoes.

Case II: Dominoes that show exactly two different numbers (84 dominoes)
There are two different ways to have two different numbers appear.

(i) Outside values equal, middle value different, such as 3-6-3 or 2-0-2. There are P(7,2) = 42 of these, formed by choosing one of 7 values for the middle spot and one of 6 for the other two. There is no duplication here, because we've only filled two spots; the third is determined by one of the first two.

(ii) Two equal values next to each other, such as 1-1-2 or 4-5-5. We count only dominoes of the type x-x-y here, for the domino y-x-x is the same. There are 7 ways to choose the value that appears twice on this dominoe and 6 ways to choose the single value. There are P(7,2) = 42 such dominoes.

Case III: Dominoes that show exactly three different numbers (105 dominoes)
These are P(7,3) = 210 of these, counted by choosing from among 7 values for the first spot, from among 6 for the second, and from among the remaining 5 for the lasy spot. However, this technique counts the same domino twice, for the domino 3-2-6, for instance, is also counted as 6-2-3, but this is actually only one domino. We divide P(7,3) by 2 to get the actual number of different dominoes of this type. There are 105 of these.