FOUR PRACTICE PROBLEMS--PERMUTATIONS AND …



FOUR PRACTICE PROBLEMS--PERMUTATIONS AND COMBINATIONS

1. Permutations/Combinations--Executive Decision

A company president is deciding how to fill three vice-presidencies in the company: VP-Marketing, VP-Finance, and VP-Production. Twelve executives are eligible and qualified for promotion, and each could fill any of the three positions. In how many ways can the positions be filled?

See if you can do the problem two (slightly) different ways, which might actually represent two different methods of thinking that the decision-maker might use.

2. Imaginary Lottery Game

The Lottery Commission is considering a new game in which five balls would be withdrawn from a box containing 10 balls, numbered 0 to 9. The five balls would come out of the box at nearly the same time, as they do in the current Lotto game, in which six balls come out of a box into a tube at nearly the same time.

In this new game, however, the winning ticket must have the five lucky numbers in the same order as they came out of the box.

What is the chance of winning with a single five-number ticket?

3. Permutations/Combinations--Dinner Party

a. "I forgot to buy vegetables for our dinner party tonight. Will you go back to the store and get three bags of frozen vegetables?" If the store has 10 bags each of 20 different frozen vegetables, in how many ways, with respect to kinds of vegetables, can the errand be performed?

b. "I forgot to buy vegetables for our dinner party tonight. Will you go back to the store and get three bags of different frozen vegetables?" If the store has 10 bags each of 20 different frozen vegetables, in how many ways, with respect to kinds of vegetables, can the errand be performed?

c. In setting the dinner table, including place cards, for eight people, how many seating arrangements are possible?

d. What if it is a round table and it really does not matter exactly where people sit, but is does matter who is sitting next to whom. How many seating arrangements are possible with 8 people?

4. Card Game

a. If you are dealt a hand of five cards from a standard deck of 52 cards, what is the probability that your hand contains four aces? (This can be done using a permutation/combination computation, but there is another way.)

b. What is the probability that your hand contains a straight flush--a sequence of five consecutive cards, all of the same suit? (A-K-Q-J-10 is the highest sequence in each suit, and 6-5-4-3-2 is the lowest.)

SOLUTIONS

1. Permutations/Combinations--Executive Decision

Duplicates are not possible.

Method One: The decision-maker first selects three people from among the twelve, not yet thinking about their job assignments (order not important). This can be done C(12,3) = 220 ways. Then the decision-maker assigns the three chosen people to the three jobs (order important). This can be done P(3,3) = 6 ways. So the total number of ways is 220 x 6 = 1,320.

Method Two: The decision maker selects a person and immediately assigns him/her to one of the three positions (order important). This is repeated two more times. The number of ways is P(12,3) = 1,320.

Method Three--formula-free sequential method: 12 people can be selected for the first position, 11 for the second, and 10 for the third. 12 x 11 x 10 = 1,320.

2. Imaginary Lottery Game

Order is important, duplicates are not possible. P(10,5) = 30,240.

Formula-free sequential method: the first number has 10 possibilities, the second number has 9, the third number has 8, the fourth number has 7, and the fifth number has 6. 10 x 9 x 8 x 7 x 6 = 30,240.

So the probability of winning is 1 / 30,240 = 0.000033069.

3. Permutations/Combinations--Dinner Party

a. Order is not important, duplicates are possible. C-bar (20,3) = 1,540.

Note: As long as there are three or more bags of each vegetable, the actual number of bags of each vegetable is not relevant. The question asked how many ways "with respect to kinds of vegetables" can the errand be performed. Therefore "n" is the number of kinds of vegetables, not the total number of bags.

b. Order is not important, duplicates are not possible. C(20,3) = 1,140.

c. Order is important, duplicates are not possible. P(8,8) = 40,320.

Formula-free: going around the table, there are 8 choices for the first seat, 7 choices for the second seat, etc. 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320.

d. The first person can sit anywhere. There are then 7 choices for the person to the first person's right, 6 choices for the person to that person's right, etc. 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040. This is also equal to P(7,7).

4. Card Game

a. Method One: There are C(52,5) = 2,598,960 possible hands. 48 of these contain 4 aces (4 aces and the king of spades, 4 aces and the queen of spades, 4 aces and the jack of spades, etc.).

48 / 2,598,960 = 0.000018469, or one chance in 54,145.

Or, the "sequential approach." As you pick up your cards, what is the probability of picking up four aces, followed by a non-ace? By the multiplicative rule, this is:

4/52 x 3/51 x 2/50 x 1/49 x 48/48 = 0.0000036937852.

But you do not have to get the four aces first and the non-ace last. You could get the non-ace fourth:

4/52 x 3/51 x 2/50 x 48/49 x 1/48 = 0.0000036937852.

Or you could get the non-ace third:

4/52 x 3/51 x 48/50 x 2/49 x 1/48 = 0.0000036937852.

Or you could get the non-ace second:

4/52 x 48/51 x 3/50 x 2/49 x 1/48 = 0.0000036937852.

Or you could get the non-ace first, followed by the four aces:

48/52 x 4/51 x 3/50 x 2/49 x 1/48 = 0.0000036937852.

By the addition rule, the union of all of the "or's" above would be the sum of the five individual probabilities, which is 0.000018469, in exact agreement with the first method.

b. There are C(52,5) = 2,598,960 possible hands. Among these, how many are "straight flushes?" There are nine straight-flushes in each suit: A-K-Q-J-10, K-Q-J-10-9, Q-J-10-9-8, J-10-9-8-7, 10-9-8-7-6, 9-8-7-6-5, 8-7-6-5-4-, 7-6-5-4-3, and 6-5-4-3-2. There are four suits, so there are 36 possible straight flushes.

36 / 2,598,960 = 0.000013852 or one chance in 72,193.

Note that this is rarer than four-of-a-kind, even four aces. So, in poker, any straight flush will beat any four-of-a-kind.

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