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Working with whole numbers only . . .
- The smallest one is 100,000,000 .
- The largest one is 999,999,999 .
That's (all the counting numbers up to 999,999,999) minus (the first 99,999,999 of them) =
900,000,000 numbers (nine hundred million)
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∙ 11y ago
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How many 8 digit numbers r there?
Thr r nnt mlln ght dgt nmbrs.
What 3 digit number multiplied by a 2 digit number equals a four digit number and you must use all numbers 1 through 9?
I think you r cheating Mr.Cheater
How many seven digits number can be made from numbers 123456789 but no repititions are allowed?
181440 possible numbers. 9 choices for first digit, leaving 8 choices for the second digit for each of these choices for the first digit, leaving 7 choices for the third digit for each of these choices for the second digit, leaving ... 3 choices for the seventh digit for each of these choices for the sixth digit, giving 9 x 8 x 7 x ... x 3 = 181440 possible numbers. More generally, when there are n different items and r need to be selected from them in order it is called a permutation and the number of ways of doing this is: nPr = n!/(n-r)! where the exclamation mark means "factorial" which is the product of all numbers from 1 to the number, that is n! = n x (n-1) x (n-) x ... x 2 x 1. In this case, there are n=9 items and r=7 need to be selected giving: 9P3 = 9!/2! = 9 x 8 x ... x 3 = 18144
If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
How many combinations of 9 if you had to use all 9?
Applying the formula n!/(n - r)! to the numbers n = 9 and r = 9, where n is the number of numbers to choose from, and r is the number of numbers chosen, 9! / (9 - 9)! is equal to 9 factorial. This is equal to 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362880 combinations.
How many 8 digit numbers r there?
Thr r nnt mlln ght dgt nmbrs.
What 3 digit number multiplied by a 2 digit number equals a four digit number and you must use all numbers 1 through 9?
I think you r cheating Mr.Cheater
How many seven digits number can be made from numbers 123456789 but no repititions are allowed?
181440 possible numbers. 9 choices for first digit, leaving 8 choices for the second digit for each of these choices for the first digit, leaving 7 choices for the third digit for each of these choices for the second digit, leaving ... 3 choices for the seventh digit for each of these choices for the sixth digit, giving 9 x 8 x 7 x ... x 3 = 181440 possible numbers. More generally, when there are n different items and r need to be selected from them in order it is called a permutation and the number of ways of doing this is: nPr = n!/(n-r)! where the exclamation mark means "factorial" which is the product of all numbers from 1 to the number, that is n! = n x (n-1) x (n-) x ... x 2 x 1. In this case, there are n=9 items and r=7 need to be selected giving: 9P3 = 9!/2! = 9 x 8 x ... x 3 = 18144
If using numbers 1 thru 49 how many 6 digit combinations can be made?
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
How many different two digit numbers can you form using the digits 1 2 5 7 8 and 9 without repetition?
36 two digit numbers can be formed...(:From Rafaelrz: The question can be stated as;how many permutations of two different digits can beobtained from a set of six different digits ?Answer:nPr equals n!/(n-r) ...... for n = 6, r = 26P2 equals 6!/(6-2)! equals 30 Permutations.
How many combinations of 9 if you had to use all 9?
Applying the formula n!/(n - r)! to the numbers n = 9 and r = 9, where n is the number of numbers to choose from, and r is the number of numbers chosen, 9! / (9 - 9)! is equal to 9 factorial. This is equal to 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362880 combinations.
How many 4 digit combinations can be made with numbers 3 7 9 4?
The number of combinations possible for taking a specified sub-set of numbers, r, from a set, n, isC(n,r) = n!/[r!(n-r)!]In this case, n = 4 and r = 1, soC(n,r) = 4!/1!3!C(n,r) = 24/6C(n,r) = 4Therefore, there are four possible combinations of the numbers 3, 7, 9 and 4.==================================This contributor strongly disagrees, but is leaving the original answer hereso that others can shop and compare.With their commas, parentheses, and factorials, the formulas are certainly impressive.Only the conclusion is wrong.The question specified "4-digit" combinations, so 'n' and 'r' are both 4.Now, to come down off the pedestal and make it understandable as well asformally rigorous ..."Combination" really means different groups of 4 digits that you can select outof the four you gave us. There's only one of those groups.If you actually want to know how many different 4-digit numbers you can makefrom them, then what you want is called "permutations" of the four digits, andyou can think of it this way, without using 'n', 'r', or parentheses or factorials:The first digit can be any one of 4. For each of those . . .The second digit can be any one of the remaining 3. For each of those . . .The third digit can be either one of the remaining 2.So the total number of different ways to line them up is (4 x 3 x 2) = 24 different 4-digit numbers.
How do you multiply 2 digit by 2 digit using distributive property?
If PQ and RS are two 2-digit numbers, then PQ * RS = 100*P*R + 10*P*S + 10*Q*R + Q*S
How many natural numbers between 150 and 300 are divisible by 9?
150 ÷ 9 = 16 r 6 → the first natural number between 150 and 300 divisible by 9 is 9 × 17 (=154)300 ÷ 9 = 33 r 3 → the last natural number between 150 and 300 divisible by 9 is 9 × 33 (=297)→ there are 33 - 17 + 1 = 17 natural numbers between 150 and 300 divisible by 9.
Which 2 numbers r consecutive whole numbers of the square root of 89?
9 and 10
How many 6-digit numbers can be formed using the digits 23456 and 8 with repition?
Order does matter in this case so you need to use the formula for a Permutation with repetition allowed. This would be n^r where n=6 different numbers and r=6 possible places for each number to go when forming the full 6-digit number. 6^6 = 46,656 possible digits
Possible four digit numbers using 0-9?
If numbers can be repeated and zeroes are allowed to lead, this is simply all natural numbers in the set {0000 - 9999}, for a total of 10,000 possible combinations. If numbers cannot be repeated, this becomes a permutation problem; out of 10 possible digits, permute four of them. This evaluates as: nPr = n! / (n-r)! 10P4 = 10! / (10-4)! 10P4 = 1 * 2 * 3 * ... * 6 * 7 * 8 * 9 * 10 / 1 * 2 * 3 * 4 * 5 * 6 10P4 = 7 * 8 * 9 * 10 10P4 = 5040 There are thus 5,040 possible combinations of four of the digits in 0-9 if any digit is not used twice and 0 is allowed to be a leading digit.
