SOLUTION: how many 5 digit counting numbers contain at least one 6

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Question 952916: how many 5 digit counting numbers contain at least one 6
Found 3 solutions by MathLover1, Sonny12345, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
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We can divide all the five digit numbers into two groups.
One group which contains all such numbers which do not have 6 at all, and the other group will be of such numbers which have at least one 6.
These two groups or sets are complementary to each other. They together make the set of all 5+digit numbers.
It is easy to calculate the number of 5 digit numbers which do not contain 6 at all. If we calculate this number and subtract it from the total number of 5 digit numbers, we will get the required answer.
The total number of 5 digit numbers => 99999+-+9999+=+90000.
Out of these, the number of 5 digit numbers which do not contain 6 at all is equal to:

9%2A9%2A9%2A9%2A9+=9%5E5=+59049 (9 digits for each placeholder)
Therefore the number of five digit numbers which contain at least one 6 is:
90000+-59049+=+highlight%2830951%29

Answer by Sonny12345(1) About Me  (Show Source):
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For this problem you have to use permutations. You can see how many permutations there are in total which is 9*10*10*10*10=90,000. This is the case since the first value can be any number from 1-9 and all the other values can be any number 0-9. The values which do not have a six in them are 8*9*9*9*9= 52,488. This is the case since the first number is any number but 6 and 0, all the other numbers are any number but 6. Thus if we subtract the two values we get the number that exactly one six in them which is 90,000-52,488 = 37,512 which is the answer.

Answer by ikleyn(52778) About Me  (Show Source):
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.
This problem is solved in 3 steps.

Step 1. 

How many are there 5 digit counting numbers in all ?


The answer is  9*10*10*10*10 = 90000,  because such a number can have

    - any of  9 digits from 1 to 9 in the mostleft position;
    - any of 10 digits from 0 to 9 in the 2nd position;
    - any of 10 digits from 0 to 9 in the 3rd position;
    - any of 10 digits from 0 to 9 in the 4th position;
    - any of 10 digits from 0 to 9 in the 5th position.


In all, there are 90000 such numbers.


Step 2. 

How many are there 5 digit counting numbers that HAVE NO the digit 6 ?


The answer is 8*9*9*9*9 = 52488, because such a number can have

    - any of  8 digits from 1 to 9 except of 6 in the mostleft position;
    - any of  9 digits from 0 to 9 except of 6 in the 2nd position;
    - any of  9 digits from 0 to 9 except of 6 in the 3rd position;
    - any of  9 digits from 0 to 9 except of 6 in the 4th position;
    - any of  9 digits from 0 to 9 except of 6 in the 5th position.


Step 3. 

Now the number under the problem's question is the difference 90000 - 52488 = 37512.