2/3 of the crockery (plates) were broken, 1/2 had a broken handle, and 1/4…

For two overlapping groups A and B drawn from a common total, the size of their union is n(A ∪ B) = n(A) + n(B) − n(A ∩ B) (the inclusion-exclusion principle for two sets) — simply adding n(A) and n(B) double-counts the overlap, so it must be subtracted once. Anything belonging to neither group is the total minus this union.

1. Let the total number of crockery pieces be x.

2. Broken plates form set A, with n(A) = 2x/3; handle-broken plates form set B, with n(B) = x/2; both broken and handle-broken (the overlap) is n(A ∩ B) = x/4.

3. By inclusion-exclusion, the plates that are broken OR handle-broken (or both) number n(A ∪ B) = 2x/3 + x/2 − x/4. Writing every term over the LCM of 3, 2 and 4 (which is 12): = 8x/12 + 6x/12 − 3x/12 = 11x/12.

4. Plates with no defect at all are the total minus this union: x − 11x/12 = x/12.

5. The problem states exactly 2 defect-free plates, so x/12 = 2, giving x = 24.

Verify with x = 24: broken = 2(24)/3 = 16, handle-broken = 24/2 = 12, both = 24/4 = 6. Union = 16 + 12 − 6 = 22 defective plates, leaving 24 − 22 = 2 defect-free plates — exactly as given, confirming the total is 24.