Extra Credit: Counting And Probability

1) Evaluate. 5!

The expression "5!" represents the factorial of 5, which is calculated as the product of all positive integers from 1 to 5. In this case, 5! is equal to 5 x 4 x 3 x 2 x 1, which equals 120. Therefore, the answer is 120.

Explanation

The expression "5!" represents the factorial of 5, which is calculated as the product of all positive integers from 1 to 5. In this case, 5! is equal to 5 x 4 x 3 x 2 x 1, which equals 120. Therefore, the answer is 120.

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2) In how many ways can 6 bicycles be parked in a row?

The number of ways to park 6 bicycles in a row can be calculated using the concept of permutations. Since the order of the bicycles matters, we can use the formula for permutations of n objects taken all at a time, which is n!. In this case, there are 6 bicycles, so the number of ways to park them in a row is 6!. Evaluating 6!, which is 6 x 5 x 4 x 3 x 2 x 1, gives us the answer of 720.

Explanation

The number of ways to park 6 bicycles in a row can be calculated using the concept of permutations. Since the order of the bicycles matters, we can use the formula for permutations of n objects taken all at a time, which is n!. In this case, there are 6 bicycles, so the number of ways to park them in a row is 6!. Evaluating 6!, which is 6 x 5 x 4 x 3 x 2 x 1, gives us the answer of 720.

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3) Which word has more permutations of its letters, HAWAII OR OREGON? (Type your answer in capital letters only)

The word "OREGON" has more permutations of its letters compared to the word "HAWAII". This is because the word "OREGON" has 6 distinct letters, while the word "HAWAII" only has 4 distinct letters. The number of permutations of a word is determined by the factorial of the number of distinct letters in the word. Therefore, "OREGON" has more permutations than "HAWAII".

Explanation

The word "OREGON" has more permutations of its letters compared to the word "HAWAII". This is because the word "OREGON" has 6 distinct letters, while the word "HAWAII" only has 4 distinct letters. The number of permutations of a word is determined by the factorial of the number of distinct letters in the word. Therefore, "OREGON" has more permutations than "HAWAII".

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4) Mr. Han is going out for the evening. He will put on one of 7 suits, one pair out of 4 pairs of shoes, and go to one of 10 restaurants. In how many ways can this be done?

There are 7 options for the suit, 4 options for the shoes, and 10 options for the restaurant. To find the total number of ways, we multiply the number of options for each category: 7 x 4 x 10 = 280. Therefore, there are 280 ways Mr. Han can choose his suit, shoes, and restaurant for the evening.

Explanation

There are 7 options for the suit, 4 options for the shoes, and 10 options for the restaurant. To find the total number of ways, we multiply the number of options for each category: 7 x 4 x 10 = 280. Therefore, there are 280 ways Mr. Han can choose his suit, shoes, and restaurant for the evening.

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5) A standard deck of cards has 52 different cards. How many 3-card ordered arrangements can be made by selecting the 3 cards without replacement?

There are 52 cards in a standard deck. When selecting a card, we have 52 options. After selecting the first card, there are 51 options left for the second card, and then 50 options for the third card. To find the number of 3-card ordered arrangements, we multiply these options together: 52 * 51 * 50 = 132,600. Hence, the correct answer is 132,600.

Explanation

There are 52 cards in a standard deck. When selecting a card, we have 52 options. After selecting the first card, there are 51 options left for the second card, and then 50 options for the third card. To find the number of 3-card ordered arrangements, we multiply these options together: 52 * 51 * 50 = 132,600. Hence, the correct answer is 132,600.

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6) How many 4-letter code symbols can be formed with the letters, P, D, Q, X without repetition?

To find the number of 4-letter code symbols that can be formed without repetition, we need to use the concept of permutations. Since there are 4 available letters (P, D, Q, X) and we need to choose 4 letters without repetition, the number of possible permutations is 4! (4 factorial), which is equal to 4 x 3 x 2 x 1 = 24. Therefore, 24 4-letter code symbols can be formed with these letters.

Explanation

To find the number of 4-letter code symbols that can be formed without repetition, we need to use the concept of permutations. Since there are 4 available letters (P, D, Q, X) and we need to choose 4 letters without repetition, the number of possible permutations is 4! (4 factorial), which is equal to 4 x 3 x 2 x 1 = 24. Therefore, 24 4-letter code symbols can be formed with these letters.

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7) There are 3 permutations of the letters in DAD. For example, DDA, DAD, and ADD. Find the number of permutations of DEED.

The number of permutations of a word can be found by calculating the factorial of the number of letters in the word. In this case, the word is DEED which has 4 letters. Therefore, the number of permutations is 4 factorial, which is equal to 4 x 3 x 2 x 1 = 24.

Explanation

The number of permutations of a word can be found by calculating the factorial of the number of letters in the word. In this case, the word is DEED which has 4 letters. Therefore, the number of permutations is 4 factorial, which is equal to 4 x 3 x 2 x 1 = 24.

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8) On a test a student is to select 6 out of 10 questions, without regard to order. How many ways can this be done?

The question is asking how many ways a student can select 6 out of 10 questions without regard to order. This is a combination problem, where the order of selection does not matter. The formula to calculate the number of combinations is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items to be selected. In this case, n = 10 and r = 6. Plugging these values into the formula, we get 10! / (6!(10-6)!), which simplifies to 210. Therefore, there are 210 ways the student can select 6 out of 10 questions.

Explanation

The question is asking how many ways a student can select 6 out of 10 questions without regard to order. This is a combination problem, where the order of selection does not matter. The formula to calculate the number of combinations is nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items to be selected. In this case, n = 10 and r = 6. Plugging these values into the formula, we get 10! / (6!(10-6)!), which simplifies to 210. Therefore, there are 210 ways the student can select 6 out of 10 questions.

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9) A special classroom has 8 sets of headphones for students who have difficulty hearing. How many possible combinations of students and headphones are there if 6 students in a class need to use headphones?

The question asks for the number of possible combinations of students and headphones. Since there are 6 students and 8 sets of headphones, each student can choose from any of the 8 sets of headphones. Therefore, the total number of combinations is obtained by multiplying the number of choices for each student, which is 8, six times. This can be calculated as 8^6, which equals 20,160.

Explanation

The question asks for the number of possible combinations of students and headphones. Since there are 6 students and 8 sets of headphones, each student can choose from any of the 8 sets of headphones. Therefore, the total number of combinations is obtained by multiplying the number of choices for each student, which is 8, six times. This can be calculated as 8^6, which equals 20,160.

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10) A standard deck of cards has 52 different cards. How many 3-card ordered arrangements can be made by selecting the 3 cards with replacement?

To find the number of 3-card ordered arrangements with replacement, we need to consider that after each card is selected, it is put back into the deck, making it available for selection again. Therefore, for each of the 3 cards, there are 52 options. Since these selections are made with replacement, we multiply the options together: 52 * 52 * 52 = 140,608.

Explanation

To find the number of 3-card ordered arrangements with replacement, we need to consider that after each card is selected, it is put back into the deck, making it available for selection again. Therefore, for each of the 3 cards, there are 52 options. Since these selections are made with replacement, we multiply the options together: 52 * 52 * 52 = 140,608.

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