The problem asks to find the number of monotone functions $f: [10] \to [2]$, where a function $f$ is called monotone if for all $1 \leq i \leq j \leq 10$, we have $f(i) \leq f(j)$.

Key observation:

A monotone function from $[m]$ to $[n]$ is equivalent to a weakly increasing sequence of length $m$ with values in $[n]$. For the specific case $m = 10$ and $n = 2$, this is equivalent to finding how many ways we can assign values to the 10 elements, such that the values are either 1 or 2, and they are in weakly increasing order.

Solution:

A weakly increasing sequence from $[10] \to [2]$ can be thought of as a sequence consisting of:

• Some number of 1's followed by some number of 2's.

This is equivalent to choosing how many 1's to place in the sequence, and the rest will be 2's. The number of ways to choose how many 1's (or equivalently 2's) to place is the number of ways to choose a subset of positions to assign the value 2.

This is a combinatorial problem, specifically choosing a subset of positions for the value 2. There are $\binom{10}{k}$ ways to choose $k$ positions to place 2's, where $k$ can range from 0 to 10. Hence, the total number of monotone functions is given by the binomial coefficient:

$T(10, 2) = \binom{10}{0} + \binom{10}{1} + \cdots + \binom{10}{10} = 2^{10}$

Calculation:

$2^{10} = 1024$

Thus, the number of monotone functions $T(10, 2)$ is $\boxed{1024}$.

Would you like further clarification or details?

Here are 5 follow-up questions you may consider:

1. What is the general formula for $T(m, n)$?
2. How do monotone functions differ when $m = n$?
3. What is the significance of weakly increasing sequences in combinatorics?
4. How does the result change if $T(10, 3)$ is calculated?
5. Can you provide an example of a monotone function when $T(4, 2)$?

Tip: When working with binomial coefficients, it's helpful to remember the properties of Pascal’s triangle for quick calculations.