# SOLUTION: how many six digit numbers... a. contain 7 at least once b. start and end with a 7 c. have alternating 7 and non 7 digits

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 Click here to see ALL problems on Permutations Question 277145: how many six digit numbers... a. contain 7 at least once b. start and end with a 7 c. have alternating 7 and non 7 digitsAnswer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website!```how many six digit numbers... a. contain 7 at least once We will find the total number of 6-digit numbers and then we will subtract from that the total number of 6-digit numbers which do not contain 7 at all, then what we get will be the number of 6-digit numbers that contain 7 at lease once. 1. First we find the total number of 6-digit numbers. Take this sampole 6-digit number: 825572 We can choose a digit to go where the 8 is 9 ways (only 9 ways since we CANNOT choose 0 for the first digit). We can then choose a digit to go where the first 2 is 10 ways (we CAN choose 0). We can then choose a digit to go where the first 5 is 10 ways (we CAN choose 0). We can then choose a digit to go where the second 5 is 10 ways (we CAN choose 0). We can then choose a digit to go where the 7 is 10 ways (we CAN choose 0). We can then choose a digit to go where the last digit 2 is 10 ways (we CAN choose 0). So there are 9x10x10x10x10x10 or 900000 6-digit numbers. [Note: we could have done that part this easier way: The largest 6 digit number is 999999. That means there are 999999 positive integers from 1 to 999999, but the integers from 1 to 99999 have less than 6 digits, so to find the number of 6-digit numbers we subract 999999-99999=900000. But either way we get the same answer 900000.] 2. Now we calculate the number of 6 digit numbers that do not contain a 7 Take this sample 6-digit number which does not contain a 7 899168 We can choose a digit to go where the first 8 is 8 ways (only 8 ways since we CANNOT choose 0 or 7 for the first digit). We can then choose a digit to go where the first 9 is 9 ways (we CAN choose 0 but not 7). We can then choose a digit to go where the second 9 is 9 ways (we CAN choose 0 but not 7). We can then choose a digit to go where the 1 is 9 ways (we CAN choose 0 but not 7). We can then choose a digit to go where the 6 is 9 ways (we CAN choose 0 but not 7). We can then choose a digit to go where the last digit 8 is 9 ways (we CAN choose 0 but not 7). So there are 8x9x9x9x9x9 or 472392 6-digit numbers that do not contain 7. 3. Now we subtract 472392 from 900000 and get 427608 ------------------------ b. start and end with a 7 Take this sample 6-digit number which starts and ends with a 7 734767 We can choose a digit to go where the first 7 is 1 way (we MUST choose it to be 7). We can then choose a digit to go where the 3 is 10 ways (we CAN choose 0). We can then choose a digit to go where the 4 is 10 ways (we CAN choose 0). We can then choose a digit to go where the second 7 is 10 ways (we CAN choose 0). We can then choose a digit to go where the 6 is 10 ways (we CAN choose 0). We can choose a digit to go where the last 7 is 1 way (we MUST choose it to be 7). So there are 1x10x10x10x10x1 or 10000 6-digit numbers that do not contain 7. c. have alternating 7 and non 7 digits. Ther can be like this: 747379 where the 7's go 1st, 3rd and 5th The number of way is 1x9x1x9x1x9 or 729 or like this: 579717 where the 7's go 2nd 4th and 6th. The number of way is 8x1x9x1x9x1 or 648 Answer = 729+648=1377 Edwin```