Question 15

We need to form 4-digit numbers $d_1 d_2 d_3 d_4$ where each $d_i \in \{1, 2, 3\}$ and $d_1 \le d_2 \le d_3 \le d_4$.

Because the digits must be in non-decreasing order from left to right, once any multiset of 4 digits is chosen from $\{1, 2, 3\}$, there is exactly one way to arrange them in non-decreasing order.

Thus, the number of such 4-digit numbers equals the number of ways to choose 4 elements from a set of 3 distinct elements with replacement (combinations with repetition).

Using the stars and bars formula $\binom{n + r - 1}{r}$ with $n = 3$ and $r = 4$:

Direct enumeration verifies the result:

• All digits identical: $1111, 2222, 3333$ (3 numbers)
• Three of one digit, one of another: $1112, 1113, 1222, 2223, 1333, 2333$ (6 numbers)
• Two of one digit, two of another: $1122, 1133, 2233$ (3 numbers)
• Two of one digit, one each of other two: $1123, 1223, 1233$ (3 numbers)
  Total $= 3 + 6 + 3 + 3 = 15$.