Combinations

What we see: a Problem Solving counting question. There are \(X\) families of \(Y\) people each, and every family must stand together in the row.

Step 1. Order the families as blocks.
What we see: each family stays together, so in the row each family acts like one block.
What it means: we are first arranging \(X\) blocks.
What we do and why: \(X\) blocks can be lined up in \(X!\) ways.

Step 2. Order the people inside each block.
What we see: inside one family, the \(Y\) members can stand in any order.
What it means: each family has \(Y!\) orders of its own, and each family picks its order on its own, separate from the others.
What we do and why: we multiply \(Y!\) once for every family, so \(Y! \times Y! \times \dots\) (\(X\) times) \(= (Y!)^X\).

Step 3. Put it together.
Every block order can pair with every inside order, so we multiply: \(X! \times (Y!)^X\).

The trap: A, \(X! \times Y!\). It counts the inside order for only one family, as if all families had to use the same order.
Faster way: test small numbers. 2 families of 2 people: blocks \(2! = 2\) ways, each family 2 ways, so \(2 \times 2 \times 2 = 8\). \(X! \times (Y!)^X\) gives \(2! \times (2!)^2 = 8\). A gives 4.
Takeaway: "must stay together" means: arrange the groups, then multiply by the inside arrangements of every group.

Answer: D