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Explanation The question asks to evaluate 6!. The exclamation mark denotes the factorial function, which means multiplying a number by all the positive integers less than it. Therefore, 6! is calculated as 6 x 5 x 4 x 3 x 2 x 1, which equals 720.
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2.
Evaluate P(7, 3)
Explanation The expression P(7, 3) represents the permutation of selecting 3 items from a set of 7 items in a specific order. The formula for permutations is P(n, r) = n! / (n-r)!, where n is the number of items in the set and r is the number of items being selected. In this case, P(7, 3) = 7! / (7-3)! = 7! / 4! = (7 * 6 * 5 * 4!) / 4! = 7 * 6 * 5 = 210. Therefore, the correct answer is 210.
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3.
How many ways can four people be arranged in a straight line?
Explanation There are 4 people to be arranged in a straight line. The first person can be chosen in 4 ways, the second person can be chosen in 3 ways (as one person is already placed), the third person can be chosen in 2 ways (as two people are already placed), and the fourth person can be chosen in 1 way (as three people are already placed). Therefore, the total number of ways to arrange the four people is 4 x 3 x 2 x 1 = 24.
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4.
How many different arrangements are there of the letters in the word ZAMBONI ?
Explanation The word "ZAMBONI" has 7 letters. To find the number of different arrangements, we use the formula for permutations of a set, which is n!, where n is the number of elements in the set. In this case, n = 7. Therefore, the number of different arrangements of the letters in the word "ZAMBONI" is 7! = 5040.
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5.
The number of distinct permutations of the letters in the word "MASSIVE" is ___________.
Explanation To find the number of distinct permutations of the word "MASSIVE," we first count the total number of letters, which is 7. The formula for calculating permutations when there are repeated letters is:
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6.
5 people are lined up for a 1000 meter race. Determine the number of different ways the first and second positions can be assigned.
Explanation There are 5 people lined up for the race, and we need to determine the number of different ways the first and second positions can be assigned. Since the order matters, we can use the permutation formula. The number of ways to choose the first position is 5, and once the first position is assigned, there are 4 people remaining for the second position. Therefore, the total number of different ways is 5 * 4 = 20.
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7.
You disconnect the thermostat from your furnace but neglect to jot down how the wires connect to it. There are three wires coming from the wall (red, black, and white) but four terminals on the thermostat. How many different ways can the wires be connected?
Explanation There are four terminals on the thermostat, but only three wires coming from the wall. This means that one of the terminals will not be connected to any wire. The wire can be connected to any of the four terminals, so there are four possible options for the first wire. After the first wire is connected, there are three remaining terminals for the second wire to be connected to. Finally, there are two remaining terminals for the third wire to be connected to. Therefore, the total number of different ways the wires can be connected is 4 x 3 x 2 = 24.
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8.
How many arrangements are there of the letters in FERMAT if you use all the letters but each arrangement must end in a vowel?
Explanation There are 6 letters in the word FERMAT. Out of these, 3 letters (E, A, and T) are vowels. In order for an arrangement to end in a vowel, the last letter must be one of these 3 vowels. Therefore, there are 3 options for the last letter. The remaining 5 letters can be arranged in any order. Using the formula for permutations, we can calculate the total number of arrangements as 3 options for the last letter multiplied by the number of permutations of the remaining 5 letters (5!). This gives us a total of 3 * 5! = 3 * 120 = 360. However, since the question specifies that each arrangement must end in a vowel, we need to subtract the number of arrangements where the last letter is not a vowel. There are 3 consonants in the word FERMAT, and the remaining 4 letters can be arranged in any order. Therefore, there are 3 * 4! = 3 * 24 = 72 arrangements where the last letter is not a vowel. Subtracting this from the total number of arrangements, we get 360 - 72 = 288. However, this calculation includes arrangements where the first letter is a vowel, which is not allowed according to the question. There are 2 vowels (E and A) that can be in the first position, and the remaining 4 letters can be arranged in any order. Therefore, there are 2 * 4! = 2 * 24 = 48 arrangements where the first letter is a vowel. Subtracting this from the total number of arrangements, we get 288 - 48 = 240. So, there are 240 arrangements of the letters in FERMAT if each arrangement must end in a vowel.
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9.
For security, an e-mail program requires you to have a ten-character password adn change it every three weeks. You like your current password, XXQWERTYZZ, because it is relatively easy to remember and type in. At the very least, you would like to keep the same letters when you change your password, How many different passwords can be created using your letters while keeping QWERTY together.
Explanation Consider QWERTY as a single item. 5!/[(2!)(2!)]
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10.
A basketball team is lining up for a team photo. There are eight players and two coaches. How many ways can they be arranged in a straight line if the two coaches must be beside each other?
Explanation 9! x 2! (consider the two coaches as a single item and then consider the different arrangements of the two coaches after)
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