CAT 2022 Slot 3 · Quant · Question 8

The digits are 1, 4, 2 and 1, with the digit 1 repeated.

The number of distinct arrangements is \frac{4!}{2!}=12, and none begins with a zero since there is no zero.

By symmetry each digit appears in the units place as often as in any other place, so count the appearances once.

Fixing a 1 in a given place leaves the distinct digits 1, 2 and 4, which arrange in 3!=6 ways.

Fixing the 2 there leaves 1, 1 and 4, giving \frac{3!}{2!}=3 ways.

Fixing the 4 there leaves 1, 1 and 2, again giving 3 ways.

Those add to 6+3+3=12, confirming the count.

The digit total in any one place is 1\times 6+2\times 3+4\times 3=6+6+12=24.

Since this holds for all four places, the sum of all twelve numbers is 24\times 1111=26664.

The mean is \frac{26664}{12}=2222 — the second option.

Shortcut. The mean of all rearrangements is the average digit repeated in every place. Here the digits average \frac{1+4+2+1}{4}=2, so the mean is 2222 — no counting needed at all.