How many 7-digit telephone numbers have an odd number of even numbers? ((7 choose 1)*5^7) + (7 choose 3)*5^7) + (7 choose 3)*5^7) + (7 choose 1)*5^7)
This is how I attempted to solve the problem, but I'm not sure if its correct.
 A: Assuming all $7$-digit numbers are admissible, including those with leading zeroes, note that
$$N \mapsto 9999999 - N$$
is a bijection from the set of $7$-digit numbers to itself with the property that it interchanges even digits with odd. So, it maps a number with $n$ even digits to a number with $7 - n$ even digits, and thus a number with an odd number of even digits to one with an even number.
This gives a bijection between the $7$-digits numbers with an odd number of even digits and those with an even number, so the counts of those two types are the same. Since there are $10^7$ $7$-digit numbers, there are
$$\tfrac{1}{2} \cdot 10^7 = 5 \cdot 10^6$$
$7$-digits numbers with an odd number of even digits.
A: We assume that all strings of $7$ digits are allowed.
Take any string of $6$ digits. If it has an odd number of even digits, it can be completed to a number with an odd number of even digits by appending any of the $5$ odd digits.
If it has an even number of even digits, it can be completed to a number with an odd number of even digits by appending any of the $5$ even digits.
So any of the $10^6$ $6$-digit strings can be completed in $5$ ways to a $7$-digit string with an odd number of even digits. 
It follows that the total number is $5\times 10^6$. 
Remark: I prefer to say that half of all the $10^7$ possible $7$-digit telephone numbers qualify.
Let us suppose that there are restrictions on the kind of telephone numbers available, but that for any allowed $6$-digit initial string can be completed to a legitimate telephone number in $10$ ways. Then our argument goes through with essentially no change, and we find that if there are $N$ legal telephone numbers, then $\frac{N}{2}$ of them have an odd number of even digits.  
