What is the number of permutation of the digits 1, 2, 3, 4 and 5 so that the sum of the digits at the first two places is smaller than the sum of the digit at the last two places?

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Official Explanation

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The number of ways the sum of the first two digits exceeds that of the last two is the same as the number of ways the last two exceed the first two.

The total number of permutations is:

5! = 120

Now need to identify patterns where the sum of the first two EQUALS sum of the last.

Possible equal-sum pairs: (1,5) and (2,4); (2,3) and (1,4); (4,3) and (5,2).

The digits in each pair can be arranged 2 ways, and the pairs themselves can be swapped (2 ways), giving:

2 \times 2 \times 2 = 8

ways per pattern.…

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What this question tests

What it tests

Whether you can count arrangements and selections correctly — permutations vs. combinations.

Common trap

Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.