A wholesaler has 12 identical steel tumblers to send out to 3 different retail shops. In how many ways can the consignment be split up, given that a shop may be sent none at all?

The tumblers are identical, so only the three counts matter — the split is a solution of a + b + c = 12 in whole numbers, zero allowed. Lining up 12 tumblers and inserting 2 dividers among the 14 positions gives C(12 + 3 − 1, 3 − 1) = C(14,2) = 91. Trap 55 = C(11,2) is the count when every shop must get at least one, which the question expressly allows against. Trap 78 = C(13,2) is an off-by-one slip in the divider count. Trap 364 = C(14,3) picks the wrong index — the dividers number 2, not 3.