Question 1

Find the coefficient a of the term in the expansion of the binomial. Binomial Term

Accepted Answer

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Question 2

Find the number of distinguishable permutations of the group of letters.

Accepted Answer

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Question 3

Decide which of the scenarios below should be counted using permutations or combinations. Scenario I:
Number of ways 7 people can line up in a row for concert tickets.
Scenario II:
Number of ways a team of 8 students can be selected from a group of 50 people.
Scenario III:
Number of ways a jury of 11 people can be selected from a group of 40 people.

Accepted Answer

Scenario I involves arranging a group of people in a specific order, so it is a permutation. Scenarios II and III involve selecting members from a group without regard to order, so they are combinations. Therefore, choice A is the correct answer.

Question 4

At a high school cafeteria, diners can choose one vegetable from a choice of 4 vegetables, one meat from a choice of 3 meats, one serving of bread from among 4 breads, and a dessert from among 3 desserts.How many meal configurations are possible?

Accepted Answer

The total number of meal configurations is calculated by multiplying the number of options for each category together: 4 (vegetables) * 3 (meats) * 4 (breads) * 3 (desserts) = 144.

Question 5

Find the specified nth term in the expansion of the binomial.

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Question 6

Twenty-four students, of whom four are seniors, are selected as semifinalists for a literary award.Of the twenty-four students, twelve finalists will be selected.In how many ways can the twelve finalists contain two seniors?

Accepted Answer

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Question 7

In how many ways can a 12-question true-false exam be answered? (Assume that no questions are omitted.)

Accepted Answer

Each question has 2 possible answers (true or false), and there are 12 questions. Therefore, the total number of ways to answer the exam is $2^{12} = 4096$.

Question 8

A combination lock will open when the right choice of three numbers (from 1 to 36) is selected.How many different lock combinations are possible?

Accepted Answer

To find the total number of possible lock combinations, we need to calculate 36 Choose 3 (denoted as C(36,3)) which is equal to:

C(36,3) = (36!)/(3!(36-3)!)

Simplifying this expression, we get:

C(36,3) = (36 x 35 x 34)/(3 x 2 x 1)

C(36,3) = 12 x 35 x 34

C(36,3) = 14, 616

Therefore, the total number of lock combinations possible is 14,616.

Question 9

Two six-sided dice are tossed.Find the probability that the sum is prime or even.

Accepted Answer

We can find the number of ways to get a prime or even sum by computing the complementary probability (i.e. the probability of rolling an odd non-prime sum), and then subtracting from 1.

To get an odd non-prime sum, we can either roll an odd number on the first die (3 ways), and a non-prime even number on the second die (2 ways), or an even number on the first die (3 ways), and a non-prime odd number on the second die (3 ways).

So there are a total of 3 x 2 + 3 x 3 = 15 ways to get an odd non-prime sum out of 6 x 6 = 36 possible outcomes when rolling two dice.

Therefore, the probability of getting a prime or even sum is 1 - 15/36 = 7/12, which corresponds to choice B.

Question 10

In a certain state, each automobile license plate number consists of two letters followed by a five-digit number.To avoid confusion between "O" and "zero" and "I" and "one", the letters "O" and "I" are not used.How many distinct license plate numbers can be formed?

Accepted Answer

There are 24 letters in the alphabet excluding I and O.
Thus, there are 24 choices for the first letter and 23 for the second letter.
So, there are $24*23=552$ ways to choose the two letters.
For each choice of letters, there are $10^5=100,000$ ways to choose a five-digit number.
Therefore, the total number of distinct license plate numbers is $552*100,000=55,200,000$.
So the answer is B.

Question 11

Evaluate the expression.

Accepted Answer

The answer of Evaluate the expression. 
...

Question 12

There are 13 patients in Dr.Ziglar's waiting room.Dr.Ziglar can see 6 patients before lunch.In how many different orders can Dr.Ziglar see 6 of the patients before lunch?

Accepted Answer

This is a permutations problem, where the order in which the patients are seen matters. The number of permutations of 6 patients from 13 is 13P6, which can be calculated as:
13P6 = 13!/7! = 13 x 12 x 11 x 10 x 9 x 8 = 1,235,520
Therefore, the answer is E.

Question 13

Twenty-four students are selected as semifinalists for a literary award.Of the twenty-four students, fourteen finalists will be selected.In how many ways can fourteen finalists be selected from the twenty-four students?

Accepted Answer

The answer of Twenty-four students are selected as semifinalists for...

Question 14

Evaluate: ₉P₆

Accepted Answer

The permutation formula is nPr = n! / (n-r)!. For 9P6, it is 9! / (9-6)! = 9! / 3! = 60,480.

Question 15

A coin is tossed three times.Find the probability of getting two heads.

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Question 16

A computer randomly generates an integer from 1 through 50.Find the probability of the event that a multiple of 8 is generated.

Accepted Answer

Multiples of 8 are 8, 16, 24, 32, 40, and 48. These are 6 out of the 50 possible integers, so the probability of generating a multiple of 8 is 6/50=3/25. The answer is B), 11ea89fe_4bec_a5fc_9fff_735e511a9b54_TB2036_11.

Question 17

A computer manufacturer offers a computer system with five different disk drives, two different monitors, and three different keyboards.How many different computer systems could a consumer purchase from this manufacturer?

Accepted Answer

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Question 18

Determine the number of ways a computer can randomly generate an integer divisible by 4 from 1 through 15.

Accepted Answer

To be divisible by 4, integers must end in either 0, 4, 8, or 12. Thus, out of the 15 integers, only 4 are divisible by 4: 4, 8, 12, and 15. Since the computer must choose only one integer, the answer is 4/15. However, this is not a whole number, so there must be some other explanation. Notice that 8 and 12 are both divisible by 4, meaning that if the computer chooses either of them, there is only one way to choose a number divisible by 4. Thus, the only two choices that work are 4 and 8, which gives us 2 ways in total. Note that the answer choices given are not in numerical order, so we choose option B, which corresponds to the answer of 2.

Question 19

Eight cards are chosen at random from a standard deck of playing cards.In how many ways can the cards be chosen if all eight cards are diamonds.

Accepted Answer

There are 13 diamonds in a deck of 52 cards, so there are 13 choose 8 ways to choose 8 diamonds, which is equal to 1287.

Question 20

A state lottery game requires a person to select nine different numbers from thirty-six numbers.The order of the selection is not important.In how many ways can this be done?

Accepted Answer

This problem involves using the combination formula. We have 36 numbers to choose from, and we want to select 9 of them. The order in which we select them does not matter, so we use the combination formula:
$${{n}\choose{k}}=\frac{n!}{k!(n-k)!}$$
Plugging in the numbers we get:
$${{36}\choose{9}}=\frac{36!}{9!(36-9)!}=\frac{36!}{9!27!}=126,843,969$$
Therefore, there are 126,843,969 ways to select nine different numbers from thirty-six numbers. The best choice is C.

Quiz 7: Sequences, Series, and Probability