24 Find the number of integers in the set $\{1,2,3, \ldots, 2009\}$ whose sum of the digits is 11 .
24 Answer: (133)
We consider 3 sets of integers:
(a) $0001 \sim 0999$. We count the number of nonnegative solutions of $a+b+c=11$. It is ${ }^{11+3-1} \mathrm{C}_{3-1}={ }^{13} \mathrm{C}_{2}=78$. Note that 11 should not be split as $0+0+11$ nor $0+1+10$. So the number of solutions is $78-3-6=69$.
(b) $1001 \sim 1999$. We count the number of nonnegative solutions of $a+b+c=10$. It is ${ }^{10+3-1} \mathrm{C}_{3-1}={ }^{12} \mathrm{C}_{2}=66$. Note that 10 should not be split as $0+0+10$. So the number of solutions is $66-3=63$.
(c) $2001 \sim 2009$. We see that only 2009 satisfies the property.
$\therefore$ The total number of integers satisfying the property is $69+63+1=133$.