Problem 7 · 2020 AMC 8
Medium
Counting & Probability
careful-countingcasework
How many integers between 2020 and 2400 have four distinct digits arranged in increasing order? (For example, 2347 is one integer.)
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Answer: C — 15 integers.
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Hint 1 of 2
“Increasing order” is a gift: once you pick which digits to use, there's only one way to arrange them. So you're choosing a set of digits, never ordering them.
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Hint 2 of 2
The first two digits are forced (2, then 3, to stay between 2020 and 2400 while increasing). So just count how many ways to pick the last two digits from {4, 5, 6, 7, 8, 9}.
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Approach: increasing ⇒ choosing a set, not arranging
- Increasing means strictly ascending, so once the four digits are chosen, the order is automatic. The thousands digit is 2 (number is 2020–2400), and the next digit must beat 2 yet keep the number ≤ 2400, so it's 3.
- That leaves the last two digits: any 2 distinct values from {4, 5, 6, 7, 8, 9}. Each pair makes exactly one valid number (smaller digit first), so just count the pairs: C(6, 2) = 15.
- Why this transfers: whenever an arrangement is forced to be increasing (or decreasing), counting collapses from permutations to combinations — you pick the values and the order takes care of itself. That's the difference between C(6,2) = 15 and the larger 6×5 = 30 of ordered pairs.
Another way — list the pairs to see C(6,2):
- Last two digits from {4,5,6,7,8,9}: pick the smaller, then a larger partner. 4 with {5,6,7,8,9}=5; 5 with {6,7,8,9}=4; 6→3; 7→2; 8→1.
- 5 + 4 + 3 + 2 + 1 = 15 — the triangular-number shape that always shows up when you count unordered pairs.
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