Finding the number of numbers such that sum of digits is at most $11$

Question

Three digit number of distinct digits are to be formed by using the digits $1,2;3,4,5,9$.

$(1)$ Finding the number of numbers such that sum of digits is at most $11$.

$(2)$ A number is selected at random. Find probability that sum of digits of selected number is at least $12$.

Try: total $3$ digit number with distinct digits and sum at most $11$ is $=$ total $3$ digit number with sum at least $12$.

We will select $1$ or both number from $5,6$.

Could someone help me to solve it? Thanks.

Answer 1

1. Digit $9$ can't be selected. Out of the five remaining digits, the combination $3,4,5$ is impossible. This leaves us $_5P_3-3! = 60 - 6 = 54$ permutations.
2. $P(\text{sum of digits} \le 11) = 54/_6P_3 = \frac{54}{120} = \frac{9}{20}$, so $P(\text{sum of digits} \ge 12) = 1 - \frac{9}{20} = \frac{11}{20}$.

Answer 2

There aren't that many combinations of digits. Further, since all of the digits in the number need to be distinct, counting all of the possible numbers is easy (multiply the number of combinations by $6$).

If the sum is at most $11$, $9$ can't be one of the digits. The only one remaining that doesn't use $9$ that has a sum more than $11$ is $(3,4,5)$. So there are $(_5C_3 - 1)\cdot 6=54$ numbers for the first part.

Given that there are $_6C_3=20$ possible combinations of digits, the probability is $(20-9)/20 = 0.55$ for picking one with the sum $\geq 12$.