SOLUTION: A seven-digit number (i.e. a whole number between 1000000 and 9999999) is selected at random. What is the probability that it contains no digits other than 6's, 7's, and/or 8's

First we count the number of 7 digit integers that contain no other
digits than 6's, 7's, or 8's.

Choose the 1st digit 3 ways, either 6, 7, or 8.
Choose the 2nd digit 3 ways, either 6, 7, or 8.
Choose the 3rd digit 3 ways, either 6, 7, or 8.
Choose the 4th digit 3 ways, either 6, 7, or 8.
Choose the 5th digit 3 ways, either 6, 7, or 8.
Choose the 6th digit 3 ways, either 6, 7, or 8.
Choose the 7th digit 3 ways, either 6, 7, or 8. 

That's 3∙3∙3∙3∙3∙3∙3 = 37 = 2187 ways

That's the numerator of the desired probability,
the number of ways we can succeed in drawing a
7 digit integers that contains no other digits 
than 6's, 7's, or 8's.

Now we count the number of ways to choose ANY
7-digit number between 1000000 and 9999999:

Two ways to do that.  We could do it as above, but
the easy way is to realize there are 9999999 integers'
from 1 through 9999999, inclusive, and we eliminate
those between 1 through 999999, those with fewer
than 7 digits, and there are 999999 of those, so we
subtract:

9999999
-999999
-------
9000000 = the denominator of the desired probability,
the number of ways to succeed or fail.

Desired probability = 2187/9000000 which reduces by
dividing top and bottom by 9:

Answer = 243/1000000

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What is the probability it contains two 6's, three 7's, and two 8's?

There are 7 positions a digit in a 7-digit integer which a digit can
occupy.

There are '7 positions choose 2' or 7C2=21 ways to fill in the two 6's.
There then remain 5 positions to fill. 
There are '5 positions choose 3' or 5C3=10 ways to fill in the three 7's.
There remain 2 positions to fill.
There are '2 positions choose 2' or 2C2=1 way to fill in the two 8's.

Answer = (7C2)(5C3)(2C2) = 21∙10∙1 = 210 ways

Notice it would not have mattered if we had filled in the 7's first:

There are '7 positions choose 3' or 7C3=35 ways to fill in the three 7's.
There then remain 4 positions to fill. 
There are '4 positions choose 2' or 4C2=6 ways to fill in the two 6's.
There remain 2 positions to fill.
There are '2 positions choose 2' or 2C2=1 way to fill in the two 8's.

Answer = (7C3)(4C2)(2C2) = 35∙6∙1 = 210 ways

That was not a necessary step, but just to show you that you can choose
them in any order.

Desired probability = 210/9000000 which reduces by
dividing top and bottom by 30:

Answer = 7/300000

Edwin