Find the total number of combinations of 5 alphabets A, B, A, B, B taking some…

Concept: when a collection contains repeated identical items, a selection is described only by how many copies of each distinct item it holds, never by the order in which they are picked. An item that is available r times therefore contributes r + 1 possibilities of its own — take 0, 1, 2, … or r copies of it — and because those choices are made independently, the multiplication principle gives the total number of selections as the product of all the (r + 1) factors. That product includes the selection that takes nothing at all, so a question that asks for some or all of the items removes exactly that one case.

Application: the supply here is the letter A available 2 times and the letter B available 3 times, so the two independent choices are how many A’s to take and how many B’s to take.

1. The number of A’s in a selection can be 0, 1 or 2, so the A-part can be settled in 2 + 1 = 3 ways.

2. The number of B’s in a selection can be 0, 1, 2 or 3, so the B-part can be settled in 3 + 1 = 4 ways.

3. The two counts are chosen independently of each other, so the number of selections, counting the one that takes no letter at all, is 3 × 4 = 12.

4. Taking some or all of the letters means at least one letter is taken, so that empty selection is discarded: 12 − 1 = 11.

Cross-check by listing every selection directly, grouped by how many A’s it uses:

The three group counts add to 3 + 4 + 4 = 11, which is what the product method gave. Counting by the size of the selection instead is a second independent check:

• One-letter selections: A, B — 2 in all.

• Two-letter selections: AA, AB, BB — 3 in all.

• Three-letter selections: AAB, ABB, BBB — 3 in all.

• Four-letter selections: AABB, ABBB — 2 in all.

• Five-letter selection: AABBB — 1 in all.

Adding these size-wise counts gives 2 + 3 + 3 + 2 + 1 = 11 again.

Total number of combinations taking some or all of the letters = 11.