a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one…

Key observations:

• Exactly one of ab, bc and ca is odd → exactly two numbers are odd and one number is even.

• abc is a multiple of 4 → because there is only one even, that even must be divisible by 4. Within 2 to 10 the possible even values divisible by 4 are 4 and 8.

• Arithmetic mean of a and b is an integer → a and b have the same parity. Combined with the first observation, a and b are the two odd numbers.

• Arithmetic mean of a, b and c is an integer → a + b + c is divisible by 3.

Enumerate the possibilities for the even number c (must be 4 or 8):

• If c = 4 (c ≡ 1 mod 3), then a + b ≡ 2 (mod 3). The available odd numbers are 3, 5, 7, 9. Valid unordered odd pairs with sum ≡ 2 mod 3 are (3,5) and (5,9), giving triplets {3,5,4} and {5,9,4}.

• If c = 8 (c ≡ 2 mod 3), then a + b ≡ 1 (mod 3). Using the odd numbers 3, 5, 7, 9, valid unordered odd pairs are (3,7) and (7,9), giving triplets {3,7,8} and {7,9,8}.

There are 2 valid triplets when the even is 4 and 2 valid triplets when the even is 8, for a total of 4 unordered triplets. Therefore the answer is 4.