In how many ways can 4 distinct books be distributed into 2 identical boxes such that no box is empty?

We are given:

• 4 distinct books
• 2 identical boxes
• No box can be empty

We are to find the number of distinct distributions of the books into the boxes, considering that the boxes are identical, so swapping contents doesn’t count as a new arrangement. Let’s consider partitions of the 4 books between 2 boxes (non-empty), and for each partition, count how many non-equivalent (i.e., unique under box identity) ways there are. Case 1: 1 book in one box, 3 in the other

• Choose 1 book to go in the smaller box: C(4, 1) = 4
• The remaining 3 go into the other box
• Since the boxes are identical, choosing (A in Box1, BCD in Box2) is the same as (BCD in Box1, A in Box2)
  ⇒ So we must divide by 2 to avoid double-counting

So number of unique distributions = 4 / 2 = 2 Case 2: 2 books in each box

• Choose any 2 books to go into one box: C(4, 2) = 6
• The remaining 2 go into the second box
• But since the boxes are identical, the pair {A,B} in Box1 and {C,D} in Box2 is the same as {C,D} in Box1 and {A,B} in Box2
  ⇒ So divide by 2: 6 / 2 = 3

Case 3: 3 books in one box, 1 in the other Same as Case 1 — symmetric
⇒ Also gives 2 unique ways Total = 2 (1–3 split) + 3 (2–2 split) + 2 (3–1 split) = 7