Three gnomes and three elves sit down in a row of six chairs. If no gnome will sit next to another ganome and no elf will sit next to another elf, in how different ways can the elves and gnomes sit?

Answer

To solve the problem of arranging three gnomes and three elves in a row of six chairs such that no two gnomes sit next to each other and no two elves sit next to each other, we can break down the solution into clear steps.

### Step-by-Step Solution:

1. **Understanding the Arrangement**:
We have a total of 6 chairs and we need to arrange 3 gnomes (G) and 3 elves (E) such that they alternate. This means that the only valid arrangements are either:
- G E G E G E (Gnomes in odd positions and Elves in even positions)
- E G E G E G (Elves in odd positions and Gnomes in even positions)

2. **Case 1: Gnomes in Odd Positions**:
- In this case, the arrangement will be G E G E G E.
- The odd positions (1, 3, 5) will be occupied by gnomes and the even positions (2, 4, 6) will be occupied by elves.
- The number of ways to arrange the 3 gnomes in the 3 odd positions is given by \(3!\) (factorial of 3).
- Similarly, the number of ways to arrange the 3 elves in the 3 even positions is also \(3!\).

Therefore, the total arrangements for this case is:
\[ 3! \times 3! = 6 \times 6 = 36 \text{ ways} \]

3. **Case 2: Elves in Odd Positions**:
- In this case, the arrangement will be E G E G E G.
- The odd positions (1, 3, 5) will be occupied by elves and the even positions (2, 4, 6) will be occupied by gnomes.
- Again, the number of ways to arrange the 3 elves in the 3 odd positions is \(3!\) and the number of ways to arrange the 3 gnomes in the 3 even positions is also \(3!\).

Thus, the total arrangements for this case is:
\[ 3! \times 3! = 6 \times 6 = 36 \text{ ways} \]

4. **Total Arrangements**:
- To find the total number of arrangements, we add the number of arrangements from both cases:
\[ 36 + 36 = 72 \text{ ways} \]

### Final Answer:
The total number of different ways the elves and gnomes can sit is **72 ways**.