Using the digits {1, 2, 3, 4, 5} without repetition, form 5-digit numbers. How many such numbers have the units (ones) digit greater than the tens digit?

Introduction / Context:
We count permutations with a simple inequality constraint involving only the last two positions (tens and units digits).

Given Data / Assumptions:

• Digits are 1–5, used once each.
• 5-digit permutations (all digits used).
• Constraint: units digit > tens digit.

Concept / Approach:
Symmetry on ordered pairs: for any unordered pair {a, b} with a ≠ b, exactly one of the two orders satisfies units > tens.

Step-by-Step Solution:
Choose which two digits occupy the last two places: C(5, 2) = 10 unordered choices; for each chosen pair, exactly 1 ordering has units > tens.Arrange the first three positions with the remaining 3 digits: 3! = 6.Total = 10 × 6 = 60.

Verification / Alternative check:
Out of all 5P2 = 20 ordered pairs for the last two positions, half satisfy the inequality; 20/2 = 10 ways for the last two, times 6 ways for the first three gives 60.

Why Other Options Are Wrong:
48 (= 2 × 4!) ignores the exact half-ordering logic; 54/72 are ad hoc counts.

Common Pitfalls:
Double-counting when choosing ordered pairs first; working with unordered pairs avoids that.

Final Answer:
60