[Combinatorics]
Problem: An integer $n$, with $100 \leq n \leq 999$, is chosen at random. What is the probability that the sum of the digits of $n$ is 24 ?

Solution: The number of integers between 100 and 999 inclusive is $999-100+1=900$.

An integer $n$ in this range has three digits, say $a, b$ and $c$, with the hundreds digit equal to $a$.

Note that $0 \leq b \leq 9$ and $0 \leq c \leq 9$ and $1 \leq a \leq 9$.

To have $a+b+c=24$, then the possible triples for $a, b, c$ in some order are $9,9,6 ; 9,8,7$; $8,8,8$. (There cannot be three 9's. If there are two 9's, the the other digit equals 6 . If there is one 9 , the second and third digits add to 15 but are both less than 9 , so must equal 8 and 7 . If there are zero 9's, the maximum for each digit is 8 , and so each digt must be 8 in order for the sum of all three to equal 24.)

If the digits are 9, 9 and 6, there are 3 arrangements: 996, 969, 699.



If the digits are 9, 8 and 7, there are 6 arrangements: 987, 978, 897, 879, 798, 789.

If the digits are 8,8 and 8 , there is only 1 arrangement: 888 .

Therefore, there are $3+6+1=10$ integers $n$ in the range 100 to 999 with the sum of the digits of $n$ equal to 24 .

The required probability equals the number of possible values of $n$ with the sum of digits equal to 24 divided by the total number of integers in the range, or $\frac{10}{900}=\frac{1}{90}$.

Final answer: $\frac{1}{90}$