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How many 5 digit numbers can be made using the set 1 2 3 5 7 8 with no repeating?
There are 6 digits in the set, giving A choice of 6 for the first digit, leaving a choice of 5 for the second digit for each choice of the first, leaving a choice of 4 for the third digit for each choice of the first two, leaving a choice of 3 for the fourth digit for each choice of the first three, leaving a choice of 2 for the fifth digit for each choice of the first four making: number of 5-digit numbers = 6 x 5 x 4 x 3 x 2 = 720 possible 5-digit numbers. This is a choice where the order matters and is called a permutation. To calculate a permutation of r items from a set of n the formula: nPr = n!/(n - r)! will do it, where n! means "n factorial" and is calculated as the multiple of all digits less than n but greater than 0, ie: n! = n x (n - 1) x (n - 2) x ... x 2 x 1 for example, 4! = 4 x 3 x 2 x 1 = 24. (0! - zero factorial - is defined to be 1, ie 0! = 1.) In this question, n = 6 (set of 6 digits) and a permutation of 5 digits (for a 5-digit number) is required, thus the number of permutations (5-digit numbers) is: 6P5 = 6!/(6 - 5)! = 6!/1! = 6 x 5 x 4 x 3 x 2 x 1/1 = 720 If the question had been how many 4-digit numbers could be made from that set of 6 digits, then the answer would have been: 6P4 = 6!/(6 - 2)! = 6 x 5 x 4 x 3 x 2 x 1/2 x 1 = 6 x 5 x 4 x 3 = 360
How many even four digit numbers can be made using 0235 and 6 where the first digit is not 0?
For the number to be even it must end with either 0, 2 or 6. As there is a limitation of the number starting with 0, consider those ending with 0 separately from those ending with 2 or 6 and add together the results Ending with 0: 1st digit has a choice of 4 possible digits; for each of these 2nd digit has a choice of one of the 3 remaining non-zero digits; for each of these 3rd digit has a choice of one of the 2 remaining non-zero digits; for each of these 4th must be 0 and only 1 choice numbers = 4 x 3 x 2 x 1 = 24 Ending with 2 or 6: The 4th digit has a choice of 2; for each of these 1st digit has a choice of 3; for each of these 2nd digit has a choice of 3; for each of these 3rd digit has a choice of 2 numbers = 2 x 3 x 3 x 2 = 36 Total ways = 24 + 36 = 60 possible numbers.
What is the prime factorization 2673?
Factor tree: 2673 3 x 891 3 x 3 x 297 3 x 3 x 3 x 99 3 x 3 x 3 x 3 x 33 3 x 3 x 3 x 3 x 3 x 11 The prime factorization of 2673 is 3 x 3 x 3 x 3 x 3 x 11 or 35 x 11.
What is 3 to the power of -9?
12
What two numbers are multiplied to make 25?
There are infinitely many possible answers. Let X be any number and let Y = 25/X. Then XY = X*(25/X) = 25. Since the choice of X was arbitrary, there are infinitely many possible solutions. Some of them are: -10*-2.5, 3*81/3, 1000*0.025
If 3-y is one factor of 6-3x-2y plus xy what is the other factor you also have the answer choices choice A is 2-x choice B is 3-x choice C is 2-y and choice D is x-2?
Trick question. Ordinarily that would factor to (x - 2)(y - 3) But if the first factor is 3 - y, the correct answer is A: 2 - x.
How many 5 digit numbers can be made using the set 1 2 3 5 7 8 with no repeating?
There are 6 digits in the set, giving A choice of 6 for the first digit, leaving a choice of 5 for the second digit for each choice of the first, leaving a choice of 4 for the third digit for each choice of the first two, leaving a choice of 3 for the fourth digit for each choice of the first three, leaving a choice of 2 for the fifth digit for each choice of the first four making: number of 5-digit numbers = 6 x 5 x 4 x 3 x 2 = 720 possible 5-digit numbers. This is a choice where the order matters and is called a permutation. To calculate a permutation of r items from a set of n the formula: nPr = n!/(n - r)! will do it, where n! means "n factorial" and is calculated as the multiple of all digits less than n but greater than 0, ie: n! = n x (n - 1) x (n - 2) x ... x 2 x 1 for example, 4! = 4 x 3 x 2 x 1 = 24. (0! - zero factorial - is defined to be 1, ie 0! = 1.) In this question, n = 6 (set of 6 digits) and a permutation of 5 digits (for a 5-digit number) is required, thus the number of permutations (5-digit numbers) is: 6P5 = 6!/(6 - 5)! = 6!/1! = 6 x 5 x 4 x 3 x 2 x 1/1 = 720 If the question had been how many 4-digit numbers could be made from that set of 6 digits, then the answer would have been: 6P4 = 6!/(6 - 2)! = 6 x 5 x 4 x 3 x 2 x 1/2 x 1 = 6 x 5 x 4 x 3 = 360
How many even four digit numbers can be made using 0235 and 6 where the first digit is not 0?
For the number to be even it must end with either 0, 2 or 6. As there is a limitation of the number starting with 0, consider those ending with 0 separately from those ending with 2 or 6 and add together the results Ending with 0: 1st digit has a choice of 4 possible digits; for each of these 2nd digit has a choice of one of the 3 remaining non-zero digits; for each of these 3rd digit has a choice of one of the 2 remaining non-zero digits; for each of these 4th must be 0 and only 1 choice numbers = 4 x 3 x 2 x 1 = 24 Ending with 2 or 6: The 4th digit has a choice of 2; for each of these 1st digit has a choice of 3; for each of these 2nd digit has a choice of 3; for each of these 3rd digit has a choice of 2 numbers = 2 x 3 x 3 x 2 = 36 Total ways = 24 + 36 = 60 possible numbers.
How many ways 3 students can sit in 3 chairs?
The first student has 3 chairs to choose from. Once he has mad the choice, the second student has 2 chairs to choose from. And the last student only has 1 choice, so 3 x 2 x 1 = 6 ways.Here are the six ways. For ease, call the students A, B, and C:ABCACBBACBCACABCBA
At the beginning of summer you have 6 books to read In how many orders can you read the book Is this a permutation or combination?
You have a choice of 6 books for the first book, leaving a choice of 5 for the second, for each of the choice of first book, and so on until a reading order list has been made. This can be done in 6 x 5 x 4 x 3 x 2 x 1 = 720 ways. This is a permutation as the order of choice matters. A combination is where the order doesn't matter, ie a group has been selected, for example: You have a reading list of 6 books, but only need to read 4 of them over the summer, borrowing them from a library; how many groups of 4 books can you choose from the reading list and take home from the library. In this case, you can select (6 x 5 x 4 x 3) ÷ (4 x 3 x 2 x 1) = 15 different groups of 4 books from your reading list of 6 (and when you have your group it can be ordered in 4 x 3 x 2 x1 = 24 different ways for reading).
Which choice shows how to find the least common multiple of 7 and 12 through prime factorization?
Since 7 and 12 have no prime factors in common, their LCM is their product. 2 x 2 x 3 = 12 2 x 2 x 3 x 7 = 84, the LCM
What is the prime factorization 2673?
Factor tree: 2673 3 x 891 3 x 3 x 297 3 x 3 x 3 x 99 3 x 3 x 3 x 3 x 33 3 x 3 x 3 x 3 x 3 x 11 The prime factorization of 2673 is 3 x 3 x 3 x 3 x 3 x 11 or 35 x 11.
What is 3 to the 12th power?
To find the answer multiply 3, twelve times. Ex: 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3= 531,441
What are the factors of 8775?
8775 = 3 x 2925 = 3 x 3 x 975 = 3 x 3 x 3 x 325 = 3 x 3 x 3 x 5 x 65 = 3 x 3 x 3 x 5 x 5 x 13
What is 3 to the power of -9?
12
How do you find the probability of guessing exactly three four choice multiple choice questions out of five?
The probability of guessing any one is 1 out of 4, or 0.25. Assume that the choices are made independently. Then, if X is the random variable which represents the number of successes (correct guesses), X is a Binomial variable with n = 5 and p = 0.25. Then Prob(X = 3) = 5C3*p^3*(1-p)^(n-3) = 10*(0.25)^3*(0.75)^2 = 0.088, approx.
What two numbers are multiplied to make 25?
There are infinitely many possible answers. Let X be any number and let Y = 25/X. Then XY = X*(25/X) = 25. Since the choice of X was arbitrary, there are infinitely many possible solutions. Some of them are: -10*-2.5, 3*81/3, 1000*0.025
