Doubt: AR-XT-017-20260705-1640-q3

Quick guess at where $272.5$ came from: that's $\dfrac{124 + 421}{2}$, the midpoint of the smallest and largest arrangements. That shortcut only works when the numbers in the set are symmetrically spread around the midpoint — these six aren't (two of them, 214 and 241, cluster in the 200s), so the true mean gets pulled below the midpoint. Let's compute it properly.

Step 1 — List the numbers. The digits $2, 4, 1$ are all distinct, so there are $3! = 6$ arrangements: $124,\; 142,\; 214,\; 241,\; 412,\; 421$

Step 2 — Count how often each digit sits in each position. With 6 numbers and 3 positions, each digit lands in the hundreds place in exactly $6 \div 3 = 2$ of the numbers, and the same holds for the tens and units places. So each of $2, 4, 1$ appears twice in each column.

Step 3 — Total the column contributions.

• Hundreds: $2 \times (2+4+1) \times 100 = 1400$
• Tens: $2 \times (2+4+1) \times 10 = 140$
• Units: $2 \times (2+4+1) \times 1 = 14$

Total $= 2 \times 7 \times (100+10+1) = 14 \times 111 = 1554$. (Sanity check by adding the six numbers by hand: you get $1554$ too.)

Step 4 — Divide by the count. $\text{mean} = \frac{1554}{6} = 259$

Notice the mean came out as a whole number even though the average digit $\tfrac{7}{3}$ is a fraction — the repunit $111 = 3 \times 37$ absorbs the $3$ cleanly.

Final answer: C (259)