Question

the number of ways to select two numbers from the set \( [1, 2, \ldots, 12] \) whose sum is divisible by 3:

Answer 1

Any natural number is either of the form 3k or 3k – 1 or 3k +1.

Sum of two numbers will be divisible by 3 if and only if either both are of the form 3k or one is of the form 3k - 1 and other is of the form 3k + 1.

This can be done in ⁴C₂ + ⁴C₁ x ⁴C₁ = 6 + 16 = 22 ways