CAT 2022 Slot 2 · Quant · Question 12

Digits may not repeat, so the numbers formed have at most six digits.

Any number above 2000 must have four, five or six digits, so count those three cases.

For four-digit numbers, the leading digit must make the number exceed 2000, so it is 2, 3, 4 or 5 — four choices.

Even the smallest such number, 2013, already clears 2000, so no further restriction is needed.

The remaining three places are filled from the five unused digits in 5\times 4\times 3=60 ways.

That gives 4\times 60=240 four-digit numbers.

For five-digit numbers, the leading digit just cannot be 0, leaving five choices.

The other four places are filled in 5\times 4\times 3\times 2=120 ways.

That gives 5\times 120=600.

For six-digit numbers, the leading digit again has five choices and the remaining five digits can be arranged in 5!=120 ways.

That gives another 5\times 120=600.

The total is 240+600+600=1440 — the first option.

Shortcut. Every five- and six-digit number automatically beats 2000, so only the four-digit case needs a leading-digit restriction. Noting that the five- and six-digit counts are identical — both 5\times 120 — halves the work.