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anonymous
 one year ago
How many sevendigit numbers can be formed from the digits 09 if each digit can be used more than once?
anonymous
 one year ago
How many sevendigit numbers can be formed from the digits 09 if each digit can be used more than once?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439939273412:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont know if that is right or not?

jchick
 one year ago
Best ResponseYou've already chosen the best response.1This depends on whether the digits of the number can be repeated.... is 1123456 allowed, or do the digits all have to be different. Another question is can there be a leading 0 (or string of 0s, if duplication is allowed) So, working through the options, if no duplication is allowed, and the leading character cannot be 0, then there are 9 choices when we pick the first digit, 9 when we pick the second, 8 when we pick the third, 7 for the 4th, 6 for the 5th, 5 for the 6th and 4 for the 7th. 9 x 9 x 8 x 7 x 6 x 5 x 4 = 544,320 If we can lead with a 0 then there are 10 ways to pick the first digit, then 9 for the second, and so on down to 4 for the 7th. 10 x 9 x 8 x 7 x 6 x 5 x 4 = 604,800 If the digits can be repeated, but we cannot start with leading 0s then there are 9 ways to pick the first, and then 10 for each of the other positions 9 x 10 x 10 x 10 x 10 x 10 x 10 = 9,000,000 If we can use a leading 0 then we can choose from 10 digits for all positions and that is 10 x 10 x 10 x 10 x 10 x 10 x 10 = 10,000,000

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1JosephDeng your drawing would be correct if repeats weren't allowed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1439939494880:dw is that correct?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1since repeated digits are allowed, you compute 10^7 the long way to do it is to multiply out 10*10*10*10*10*10*10

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Smaller example: imagine instead of a 7 digit number, we form a 2 digit number (using 0 through 9, repeats allowed) set up a table with 10 rows and 10 columns. Along the top row we have 0 through 9. Along the left side we have 0 through 9. There will be 10*10 = 100 cells in the table that represent 00 through 99, which is exactly 100 different numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what would you use: permutation or combination?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i get it now, but I want to know how to use the equation to solve

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you use neither because repeated digits are allowed

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1permutation and combination both have the condition that repeats aren't used

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1instead you'll use n^k where n is the number of choices per slot, and k is the number of slots

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1n = 10 choices per slot (0 through 9) k = 7 slots because we're forming 7 digit numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0reading the number from left to right, example 4293855 the first digit can be 19 , so there are 9 choices the second digit can be 09 so there are 10 choices the third digit can be 09 so there are 10 choices the fourth digit can be 09 so there are 10 choices ... the seventh digit can be 09 so there are 10 choices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So i get 9 x 10 x 10 x 10 x 10 x 10 x 10

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Oh right, I wasn't thinking. The first digit can't be 0
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