How Many Numbers Greater Than 1000 Can 1, 1, 2, 2, 3 Form?

Assume you have the digit cards 1, 1, 2, 2, and 3, and may use each card at most once. A number greater than 1000 can have four or five digits here. We count the distinct numbers of each length, taking care not to treat identical copies of a digit as different.

With this interpretation, the total is 60. If the instruction instead requires using all five digits in every number, only the five-digit count applies, giving 30.

Count the five-digit numbers

Using all five digits makes a five-digit number, which is automatically greater than 1000. If all five digits were different, there would be $5!$ arrangements. But swapping the two 1s does not create a new number, and neither does swapping the two 2s. Each distinct number is therefore counted $2! \times 2!$ times in the raw permutation count.

$$\frac{5!}{2!2!}=\frac{120}{4}=30$$

So there are 30 distinct five-digit numbers.

Count the four-digit numbers

A four-digit number uses four of the five available digit cards, so exactly one card is left out. Group the possibilities by the value of the omitted digit. Omitting either copy of the same digit gives the same remaining collection, so each value is considered only once.

Omitted digit value	Digits left	Distinct arrangements
1	1, 2, 2, 3	$4!/2! = 12$
2	1, 1, 2, 3	$4!/2! = 12$
3	1, 1, 2, 2	$4!/(2!2!) = 6$

For example, when a 1 is omitted, the remaining digits are 1, 2, 2, and 3. There are $4!$ arrangements if the two 2s are temporarily treated as different, but each actual arrangement is counted twice because exchanging those two 2s changes nothing. That gives $4!/2!=12$ distinct numbers.

Adding the three cases gives

$$12+12+6=30$$

Every arrangement counted is greater than 1000. The first digit cannot be zero because there is no zero among the supplied digits. In fact, any four-digit number made from these cards begins with 1, 2, or 3 and has a positive remaining part, so it is strictly larger than 1000.

Add the two lengths

The four-digit and five-digit cases cannot overlap: a number cannot have both lengths. Therefore,

$$30+30=60$$

There are 60 distinct numbers greater than 1000 when you may use either four or five of the supplied digits, without using any digit more often than it appears.

Common counting mistakes

• Counting five-digit numbers only: This is correct only if every number must use all five cards. If four-digit numbers are allowed, they must be included too.
• Counting identical digits as distinguishable: The two 1s and two 2s are equal digit values. Divide by the factorials of their multiplicities when arranging a multiset.
• Counting the same omission twice: Leaving out the first 1 or the second 1 produces the same four-digit collection. Count omission cases by digit value, not by physical card.
• Including numbers with fewer than four digits: No number with at most three digits is greater than 1000, so those lengths do not contribute.
• Treating 1000 as an available number: It cannot be formed from these digits, and in any case the question asks for numbers strictly greater than 1000.