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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
