Question 1

Evaluate the expression.
-${ } _ { 4 } \mathrm { P } _ { 4 }$

Accepted Answer

${ } _ { 4 } \mathrm { P } _ { 4 } = \frac{4!}{(4-4)!} = \frac{4!}{0!} = 4! = 4 \times 3 \times 2 \times 1 = 24.$

Question 2

Solve the problem.
-If a club consists of 9 members, how many different arrangements of president and vice-president are possible?

Accepted Answer

There are 9 choices for the president and 8 choices left for the vice-president. So, the total number of arrangements is 9 x 8 = 72. There are 9 choices for the president and then 8 choices remaining for the vice-president, giving a total of 9x8=72 possible arrangements.

Question 3

Solve the problem.
-Find the number of permutations of the letters in the word DOUBLE.

Accepted Answer

The number of permutations of n distinct objects is n! (n factorial). In this case, there are 6 distinct letters, so the number of permutations is 6! = 720. However, we have two O's and two U's, which means we are overcounting some of the permutations. Specifically, for each permutation where the two O's are in different positions, we have another permutation where they are swapped (but all other letters remain in the same order). The same is true for the two U's. Therefore, we need to divide the total number of permutations by 2! twice (once for the O's and once for the U's) to correct for the overcounting. This gives us a total of 720 / (2! * 2!) = 720 / 4 = 180 permutations. There are 6 letters in the word DOUBLE, so there are 6 choices for the first letter, 5 choices for the second letter (since one letter has already been chosen), 4 choices for the third letter, 3 choices for the fourth letter, 2 choices for the fifth letter, and only 1 choice for the last letter. Therefore, the total number of permutations is:
6 x 5 x 4 x 3 x 2 x 1 = 720
However, since the two O's and the two E's are indistinguishable, we must divide by the number of ways to arrange those letters, which is 2! for each pair. Thus, the final answer is:
720 ÷ (2! x 2!) = 120.

Question 4

Evaluate the expression.
-${ } _ { 10 } \mathrm { C } _ { 5 }$

Accepted Answer

${ } _ { 10 } \mathrm { C } _ { 5 }$ calculates the number of combinations of 10 items taken 5 at a time, which equals $\frac{10!}{5!(10-5)!} = 252$.

Question 5

Evaluate the expression.
-${ } _ { 3 } \mathrm { C } _ { 0 }$

Accepted Answer

${ } _ { 3 } \mathrm { C } _ { 0 } = 1$ because the combination formula ${n \choose k} = \frac{n!}{k!(n-k)!}$ simplifies to 1 when $k=0$, regardless of the value of $n$ (as long as $n \geq 0$).

Question 6

Solve the problem.
-D.J Bob, a disc jockey has 34 records that he wants to play during the next hour, but he has time to play only 12 of them. Write the expression used to determine how many different ways he can select a group of 12 records to
Play.

Accepted Answer

The correct expression to determine the number of ways to select a group of 12 records out of 34 is given by the combination formula, which is $\frac { 34 ! } { 22 ! \cdot 12 ! }$. This accounts for the fact that the order in which the records are selected does not matter.

Question 7

Evaluate the expression.
-${ } _ { 9 } C _ { 1 }$

Accepted Answer

${ } _ { 9 } C _ { 1 } = \frac{9!}{1!(9-1)!} = \frac{9!}{1!8!} = \frac{9 \times 8!}{8!} = 9$. This is because the combination formula simplifies to just 9 when selecting 1 item from 9.

Question 8

Evaluate the expression.
-${ } _ { 6 } \mathrm { C } _ { 6 }$

Accepted Answer

The expression ${ } _ { 6 } \mathrm { C } _ { 6 }$ represents the number of ways to choose 6 items out of 6, which is always 1, since there is only one way to choose all items.

Question 9

Solve the problem.
-A teacher decides to give 3 different prizes to 3 of 12 students in her class. In how many ways can she do so?

Accepted Answer

The teacher can choose the first prize winner from 12 students, then the second prize winner from 11 students (since one has already won a prize), and the third prize winner from 10 students (since now two have already won a prize). Therefore, the total number of ways to give out the prizes is 12x11x10=1320. The teacher can choose the first prize winner from the 12 students, then the second prize winner from the remaining 11 students, and finally the third prize winner from the remaining 10 students. Therefore, the total number of ways she can give out 3 different prizes is 12 × 11 × 10 = 1,320. Thus, the correct answer is option D.

Question 10

Evaluate the expression.
-$\frac { 12 ! } { 2 ! 3 ! 7 ! }$

Accepted Answer

The answer of Evaluate the expression.
-\(\frac { 12 ! }...

Question 11

Solve the problem.
-A college identification number consists of a letter followed by 4 digits. How many different identification numbers are possible if repetition is not permitted?

Accepted Answer

There are 26 choices for the initial letter, and since repetition is not permitted, there are 10 choices for each of the 4 digits (since there are 10 possible digits). Therefore, the total number of possible identification numbers is:

26 × 10 × 9 × 8 × 7 = 3,276,000

So, the answer is D. There are 26 letters in the alphabet and 10 digits (0-9). 
Therefore, the number of possible identification numbers is:
26 options for the first letter x 10 options for the second digit x 9 options for the third digit x 8 options for the fourth digit 
= 26 x 10 x 9 x 8  
= 20,160 
However, the problem states that repetition is not permitted, so we must use the permutation formula:
nPr = n! / (n-r)!
Where n is the number of options and r is the number of selections. 
In this case, we have 26 options for the first letter (n=26) and we are selecting one (r=1). We have 10 options for the second digit (n=10) and we are selecting one (r=1), and so on. 
Using the permutation formula, we get:
26P1 x 10P1 x 9P1 x 8P1 
= 26 x 10 x 9 x 8 
= 18,720 
Therefore, the correct answer is D) 3,276,000.

Question 12

Solve the problem.
-Four different colored flags will be placed on a pole, one beneath another. The arrangement of the colors indicates the message. How many messages are possible if 4 flags are to be selected from 11 different colored
Flags?

Accepted Answer

The number of possible messages is calculated using permutations since the order of the flags matters. With 11 flags to choose from and 4 positions to fill, the formula for permutations is P(n, r) = n! / (n-r)!, where n is the total number of items to pick from, r is the number of items to pick, n! denotes the factorial of n, and (n-r)! is the factorial of the difference between n and r. Here, n = 11 and r = 4, so P(11, 4) = 11! / (11-4)! = 11! / 7! = (11 × 10 × 9 × 8) = 7920.

Question 13

Solve the problem.
-A license plate is to consist of 6 digits followed by 2 letters. Determine the number of different license plates possible if the first and fifth digits must be odd, and repetition is not permitted.

Accepted Answer

For the first slot, there are 5 odd digits to choose from. For the fifth slot, there are 4 odd digits left to choose from (since repetition is not allowed). For the second, third, fourth, and sixth slots, there are 5, 4, 3, and 2 choices respectively (since repetition is not allowed). For the letters, there are 26 choices for each slot.

Therefore, the total number of possible license plates is:

5 * 5 * 4 * 5 * 3 * 2 * 26 * 26 = 21,840,000 There are 5 choices for the first digit (1, 3, 5, 7, 9) and 5 choices for the fifth digit (since it cannot be the same as the first). For the second, third, fourth, sixth, and seventh digits, there are 5, 8, 7, 6, and 26 choices, respectively. (For the second digit, there are 5 choices since it cannot be the same as the first digit; for the third digit, there are 8 choices since it cannot be the same as the first two digits; for the fourth digit, there are 7 choices since it cannot be the same as the first three digits; for the sixth digit, there are 6 choices since it cannot be the same as any of the previous 5 digits; for the seventh digit, there are 26 choices since it can be any letter of the alphabet.) Therefore, the total number of license plates is 5 × 5 × 8 × 7 × 6 × 26 = 21,840,000.

Question 14

Evaluate the expression.
-${ } _ { 8 } \mathrm { P } _ { 5 }$

Accepted Answer

${ } _ { 8 } \mathrm { P } _ { 5 }$ calculates the number of permutations of 8 items taken 5 at a time, which equals $8 \times 7 \times 6 \times 5 \times 4 = 6720$.

Question 15

Solve the problem.
-Mrs. Sartori, a teacher, decides to give identical prizes to 4 of the 10 students in her class. In how many ways can she do it?

Accepted Answer

Mrs. Sartori is choosing 4 students out of 10 to give prizes to, and the order in which she chooses them does not matter. This is a combination problem, calculated as 10 choose 4, which is $ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 $.

Question 16

Solve the problem.
-A secured elevator allows access by pressing the correct sequence of letters on a control panel. If the control panel contains the 26 letters of the alphabet and a five-letter code must be entered (repetition is not permitted).
How many different codes are possible?

Accepted Answer

The first letter can be any of the 26 letters, the second can be any of the remaining 25 letters, the third can be any of the remaining 24 letters, the fourth can be any of the remaining 23 letters, and the fifth can be any of the remaining 22 letters. Therefore, the total number of codes is:
26 x 25 x 24 x 23 x 22 = 7,893,600. Thus, the answer is C. There are 26 options for the first letter, 25 for the second, 24 for the third, 23 for the fourth, and 22 for the fifth. To find the total number of possible codes, we multiply these numbers together: 26 x 25 x 24 x 23 x 22 = 7,893,600. Therefore, the correct answer is C.

Question 17

Solve the problem.
-John's Ice Cream Parlor has 12 different flavors. Willie orders a banana split and has to select 4 different flavors. How many different selections are possible?

Accepted Answer

In this problem, we are selecting 4 different flavors out of 12 different flavors, which means we are choosing a combination of 4 flavors from a set of 12. We can use the formula for combinations to solve this problem:

nCr = n!/r!(n-r)!

where n is the total number of items in the set (12 flavors), r is the number we are selecting (4 flavors), and ! represents factorial (the product of all positive integers up to that number).

Plugging in the numbers, we get:

12C4 = 12!/4!(12-4)! = (12x11x10x9)/(4x3x2x1) = 495

Therefore, there are 495 different selections possible for Willie's banana split. Option B is the correct choice. This is a combination problem because the order of the flavors in the banana split does not matter. Using the combination formula, we get:
12 choose 4 = (12!)/(4!(12-4)!) = 495

Question 18

Solve the problem.
-Mr Johnston has 3 pairs of pants, 6 shirts, 11 ties, and 10 sports coats. Assuming that he must wear one of each, how many different outfits can he wear?

Accepted Answer

Mr Johnston can choose one pair of pants out of 3.
He can then choose one shirt out of 6.
He can then choose one tie out of 11.
Finally, he can choose one sports coat out of 10.
The total number of outfits he can wear is:
3 x 6 x 11 x 10 = 1980. To find the total number of outfits, we need to multiply the number of options for each item:
3 (pants) x 6 (shirts) x 11 (ties) x 10 (sports coats) = 1,980. Therefore, the best choice is D)1980.

Question 19

Evaluate the expression.
-${ } _ { 8 } \mathrm { P } _ { 0 }$

Accepted Answer

${ } _ { 8 } \mathrm { P } _ { 0 }$ is defined as the number of ways to choose 0 items from a set of 8, which is always 1 regardless of the size of the set.

Question 20

Solve the problem.
-On a History test, Keith Kepler must write an essay for 2 of the 7 questions in Part 1, and 4 of the 6 questions in Part 2. How many different combinations of questions can he answer?

Accepted Answer

The number of ways Keith can choose 2 questions from Part 1 is 7 choose 2 or 21.
The number of ways Keith can choose 4 questions from Part 2 is 6 choose 4 or 15.
Using the multiplication principle, the total number of different combinations of questions Keith can answer is 21 x 15 = 315. Therefore, the best choice is C. The number of ways Keith can choose 2 questions from Part 1 is $\binom{7}{2}=21$. The number of ways he can choose 4 questions from Part 2 is $\binom{6}{4}=15$. To find the total number of combinations, we multiply these two outcomes: $21	imes 15 = 315$ combinations. Therefore, the answer is (C) 315.

Quiz 12: Sequences, Series, and Probability

Evaluate the expression. - ${ } _ { 4 } \mathrm { P } _ { 4 }$

Solve the problem. -If a club consists of 9 members, how many different arrangements of president and vice-president are possible?

Solve the problem. -Find the number of permutations of the letters in the word DOUBLE.

Evaluate the expression. - ${ } _ { 10 } \mathrm { C } _ { 5 }$

Evaluate the expression. - ${ } _ { 3 } \mathrm { C } _ { 0 }$

Solve the problem. -D.J Bob, a disc jockey has 34 records that he wants to play during the next hour, but he has time to play only 12 of them. Write the expression used to determine how many different ways he can select a group of 12 records to Play.

Evaluate the expression. - ${ } _ { 9 } C _ { 1 }$

Evaluate the expression. - ${ } _ { 6 } \mathrm { C } _ { 6 }$

Solve the problem. -A teacher decides to give 3 different prizes to 3 of 12 students in her class. In how many ways can she do so?

Evaluate the expression. - $\frac { 12 ! } { 2 ! 3 ! 7 ! }$

Solve the problem. -A college identification number consists of a letter followed by 4 digits. How many different identification numbers are possible if repetition is not permitted?

Solve the problem. -Four different colored flags will be placed on a pole, one beneath another. The arrangement of the colors indicates the message. How many messages are possible if 4 flags are to be selected from 11 different colored Flags?

Solve the problem. -A license plate is to consist of 6 digits followed by 2 letters. Determine the number of different license plates possible if the first and fifth digits must be odd, and repetition is not permitted.

Evaluate the expression. - ${ } _ { 8 } \mathrm { P } _ { 5 }$

Solve the problem. -Mrs. Sartori, a teacher, decides to give identical prizes to 4 of the 10 students in her class. In how many ways can she do it?

Solve the problem. -A secured elevator allows access by pressing the correct sequence of letters on a control panel. If the control panel contains the 26 letters of the alphabet and a five-letter code must be entered (repetition is not permitted) . How many different codes are possible?

Solve the problem. -John's Ice Cream Parlor has 12 different flavors. Willie orders a banana split and has to select 4 different flavors. How many different selections are possible?

Solve the problem. -Mr Johnston has 3 pairs of pants, 6 shirts, 11 ties, and 10 sports coats. Assuming that he must wear one of each, how many different outfits can he wear?

Evaluate the expression. - ${ } _ { 8 } \mathrm { P } _ { 0 }$

Solve the problem. -On a History test, Keith Kepler must write an essay for 2 of the 7 questions in Part 1, and 4 of the 6 questions in Part 2. How many different combinations of questions can he answer?