Question 980481
There are 5 one digit odd numbers, so
.
C=number of combinations of n things taken r at a time
n=total number of odd digits=5: r=how many to add at one time=2
{{{C=n!/(n-r)!r!}}}
{{{C=5!/(5-2)!2!}}}
{{{C=(5*4*3!)/(3!2!)}}}
{{{C=20/2=10}}}
ANSWER: There are 10 numbers that are the sum of two positive single digit odd numbers.
.
If we include negatives, there are a total of 10 odd single digit numbers:
n=10; r=2
{{{C=10!/((10-2)!(2!))}}}
{{{C=(10*9*8!)/(8!2!)}}}
{{{C=90/2}}}
{{{C=45}}} 
ANSWER 2: Including negatives, there are 45 numbers that are the sum of two single digit numbers.