1. **Problem statement:** We want to find the number of functions $f:\{1,2,\ldots,10\} \to \{1,2,\ldots,10\}$ such that $f(f(i))=i$ for all $i \in \{1,2,\ldots,10\}$. 2. **Understanding the condition:** The condition $f(f(i))=i$ means that applying $f$ twice returns the original input. Such a function is called an involution. 3. **Key property of involutions:** Every involution can be decomposed into disjoint cycles of length 1 or 2 only. This is because if a cycle had length greater than 2, applying $f$ twice would not return to the original element. 4. **Counting involutions:** We need to count the number of ways to partition the set $\{1,2,\ldots,10\}$ into fixed points (1-cycles) and 2-cycles. 5. **Formula for number of involutions:** The number of involutions on a set of size $n$ is given by: $$ I_n = \sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{k! \, 2^k \, (n-2k)!} $$ where $k$ is the number of 2-cycles. 6. **Applying the formula for $n=10$:** Calculate $$ I_{10} = \sum_{k=0}^5 \frac{10!}{k! \, 2^k \, (10-2k)!} $$ 7. **Calculate each term:** - For $k=0$: $\frac{10!}{0! \, 2^0 \, 10!} = 1$ - For $k=1$: $\frac{10!}{1! \, 2^1 \, 8!} = \frac{10 \times 9}{2} = 45$ - For $k=2$: $\frac{10!}{2! \, 2^2 \, 6!} = \frac{10 \times 9 \times 8 \times 7}{2 \times 4} = 630$ - For $k=3$: $\frac{10!}{3! \, 2^3 \, 4!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{6 \times 8 \times 24} = 3150$ - For $k=4$: $\frac{10!}{4! \, 2^4 \, 2!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3}{24 \times 16 \times 2} = 4725$ - For $k=5$: $\frac{10!}{5! \, 2^5 \, 0!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{120 \times 32} = 945$ 8. **Sum all terms:** $$ I_{10} = 1 + 45 + 630 + 3150 + 4725 + 945 = 9496 $$ **Final answer:** There are $\boxed{9496}$ such functions.