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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
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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
