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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 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
How many 4 number combinations can be made from the numbers 3 and 6?
16. There are two choices for the first digit (3 and 6). For either choice of first digit there are two choices for the second digit, and so on. 2 x 2 x 2 x 2 = 16.
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
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 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 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
3 letter word that starts with X?
X ed- -verb (used with object), x-ed or x'd  /ɛkst/ Show Spelled Pronunciation [ekst] Show IPA , x-ing or x'ing  /ˈɛksɪŋ/ Show Spelled Pronunciation [ek-sing] Show IPA . 1. to cross out or mark with or as if with an x (often fol. by out): to x out an error. 2. to indicate choice, as on a ballot or examination (often fol. by in): to x in the candidate of your choice.
How many 4 number combinations can be made from the numbers 3 and 6?
16. There are two choices for the first digit (3 and 6). For either choice of first digit there are two choices for the second digit, and so on. 2 x 2 x 2 x 2 = 16.
What is the Choice in Dying Act?
x ray
