Question 588171
A four-digit number is to be formed using the digits 2,3,4,5,6,7, and 8 with no
digit used more than once in a number.  What is the probability that the number
so formed with be even or more than 7000?
<pre>
The probability has a numerator and a denominator

We will calculate the numerator first, the number of successful cases and save
the denominator till later (for "dessert") as it will be a cinch to calculate
compared with the numerator.  

We break them into three categories:

1. The ones less than 7000 that end in 8.
2. The ones less than 7000 that do not end in 8, (but instead end with 2, 4, 
   or 6).
3. The ones more than 7000.

----------------------------

1.  To be even and not more than 7000, and end in 8, 

We can choose the last (4th) digit just 1 way, as 8.
We can then choose the 1st digit as any of the digits 2,3,4,5, or 6, 
which is 5 ways.
We can then choose the 2nd digit as any of the 5 digits which we didn't 
choose for the 4th or 1st digit.

We can then choose the 3rd digit as any of the 4 digits we didn't 
choose for the 4th, 1st, or 2nd digits.

That's a total of 1·5·5·4 = 100 ways.

2. To be even and not more than 7000, and not end in 8, (ie. end in 2,4, or 6)

We can choose the 4th digit any of 3 ways.
We can then choose the 1st digit as any of the 4 digits of {2,3,4,5,6} which
we did not choose for the 4th digit.
We can then choose the 2nd digit as any of the 5 digits we did not
choose for the 4th or 1st digit.
We can then choose the 3rd digit as any of the 4 digits we did not
choose for the 4th, 1st, or 2nd digits.

That's 3·4·5·4 = 240 ways.

3. To be more than 7000, 
we can choose the 1st digit 2 ways 7, or 8
We can then choose the 2nd digit as any of the remaining 6 digits.
We can then choose the 3rd digit as any of the remaining 5 digits.
We can then choose the 4th digit as any of the remaining 4 digits.

That's 2·6·5·4 = 240 ways.

Grand total = 100 + 240 + 240 = 580 ways.

So the numerator of the desired probability is 580

To calculate the denominator:

we can choose the 1st digit any of 7 ways
We can then choose the 2nd digit as any of the remaining 6 digits.
We can then choose the 3rd digit as any of the remaining 5 digits.
We can then choose the 4th digit as any of the remaining 4 digits.
That's 7·6·5·4 = 840

The desired probability is 580 ways out of 840 or {{{580/840}}}

That reduces to {{{29/42}}}

Edwin</pre>