By 
[
𝑛
]
[n]we denote the set 
{
1
,
…
,
𝑛
}
{1,…,n}. A function 
𝑓
:
[
𝑚
]
→
[
𝑛
]
f:[m]→[n] is called monotone if 
𝑓
(
𝑖
)
≤
𝑓
(
𝑗
)
f(i)≤f(j)whenever 
𝑖
<
𝑗
i<j. Let 
𝑆
(
𝑚
,
𝑛
)
S(m,n) be the number of monotone injective functions from 
[
𝑚
]
[m] to 
[
𝑛
]
[n]. What is the number of monotone injective functions from 
[
10
]
[10] to 
[
20
]
[20]?



To solve this problem, we need to find the number of monotone injective functions from \([10]\) to \([20]\). Let's break this down step by step.

### Step 1: Understanding Monotone Injective Functions
- A **monotone function** is one where \( f(i) \leq f(j) \) whenever \( i < j \). This means that the function must preserve order, i.e., if \( i < j \), then \( f(i) \leq f(j) \).
- A function is **injective** (one-to-one) if distinct elements in the domain map to distinct elements in the codomain, i.e., \( f(i) \neq f(j) \) for \( i \neq j \). Therefore, each element in the domain maps to a unique element in the codomain.

### Step 2: Combinatorial Interpretation
- Since the function is injective, the size of the domain must be less than or equal to the size of the codomain. In this case, we are mapping from \([10]\) (which has 10 elements) to \([20]\) (which has 20 elements), so the function is injective and can be described as selecting 10 distinct elements from \([20]\) and arranging them in increasing order (because the function is also monotone).

- The number of ways to choose 10 distinct elements from a set of 20 elements (without regard to order) is given by the **binomial coefficient**:

\[
S(10, 20) = \binom{20}{10}
\]

### Step 3: Calculation of \(\binom{20}{10}\)
The formula for the binomial coefficient is:

\[
\binom{20}{10} = \frac{20!}{10!(20 - 10)!} = \frac{20!}{10!10!}
\]

Now, let's compute \(\binom{20}{10}\). I'll calculate this.The number of monotone injective functions from \([10]\) to \([20]\) is \(\binom{20}{10} = 184,756\).

Would you like further details on the calculation or explanation?

Here are some related questions to think about:
1. How would the number of monotone injective functions change if the domain were \([9]\) instead of \([10]\)?
2. What is the significance of injectivity in this context, and how would the result differ if the function wasn't injective?
3. Can you compute the number of monotone functions (not necessarily injective) from \([m]\) to \([n]\)?
4. How would the solution differ if \( m > n \), making it impossible to have an injective function?
5. What role does the binomial coefficient play in combinatorics, especially in counting problems like this?

**Tip:** Binomial coefficients appear frequently in combinatorics, especially in problems related to choosing subsets or arranging objects with constraints.

By [n] we denote the set {1,…,n}. A function f:[m]→[n] is called monotone if f(i)≤f(j) whenever i<j. Let S(m,n) be the number of monotone injective functions from [m] to [n]. What is the number of monotone injective functions from [10] to [20]?

Find the number of monotone injective functions from [10] to [20] using combinatorics and the binomial coefficient. This problem explores monotone functions, injectivity, and their application in counting distinct ordered subsets. Suitable for discrete math students.

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