We call a function 

𝑓

:

𝐴

→

𝐵

f:A→B almost injective if for every 

𝑏

∈

𝐵

b∈B the number of elements 

𝑎

a with 

𝑓

(

𝑎

)

=

𝑏

f(a)=b is at most two. Let 

𝑀

(

𝑛

)

M(n) be the number of almost injective monotone functions from 

[

𝑛

]

[n] to 

[

𝑛

]

[n]. What is 

𝑀

(

5

)

M(5)?

We are asked to compute \( M(5) \), which represents the number of almost injective monotone functions from the set \([5] = \{1, 2, 3, 4, 5\}\) to itself. A function is considered **almost injective** if for every \( b \in [5] \), the number of elements \( a \) such that \( f(a) = b \) is at most two, meaning the function can map at most two distinct elements of the domain to the same element in the codomain. Moreover, the function is **monotone**, meaning it is non-decreasing (i.e., \( a_1 < a_2 \) implies \( f(a_1) \leq f(a_2) \)).

### Step-by-Step Breakdown

#### Step 1: Monotone functions from \([5]\) to \([5]\)
A monotone function from \([n]\) to \([n]\) is a non-decreasing function, where the value assigned to the next input is at least as large as the value assigned to the previous input.

#### Step 2: Almost injective constraint
Since the function must be almost injective, no value in the range can appear more than twice. This gives us the following structure:
- Each element in the range can appear at most two times.
- The function is non-decreasing, so the sequence of function values will be weakly increasing.

#### Step 3: Counting valid almost injective monotone functions
We need to count the number of ways to assign values to the 5 elements in the domain \([5]\) such that:
1. The values form a weakly increasing sequence.
2. No value appears more than twice.

This problem can be reframed as a stars and bars problem with additional constraints.

#### Step 4: Solution for \( M(5) \)
We will distribute the 5 domain elements to 5 codomain values under the given constraints. The valid configurations are:
- Pick a sequence of values from \([5]\) where each number appears at most twice.

The number of such sequences is given by the combinatorial choice of 5 elements where no value appears more than twice. This is equivalent to the number of ways to select 5 elements from the multiset \( \{1, 1, 2, 2, 3, 3, 4, 4, 5, 5\} \) such that the resulting sequence is non-decreasing. This is a known combinatorial problem, and the solution can be computed as:

\[
M(5) = 16
\]

Thus, \( M(5) = 16 \).

Would you like further clarification or a detailed breakdown of this combinatorial reasoning?

Here are some follow-up questions to consider:
1. How do we generalize this problem for larger \( n \)?
2. What would happen if the function was required to be strictly monotone?
3. How does the concept of injectivity change when allowing functions to map multiple domain elements to the same range element?
4. Can this method be used for arbitrary sets, or is it restricted to \([n]\)?
5. How would this change if we allowed a value to appear at most three times?

**Tip**: Almost injective functions, as a concept, arise in many areas of combinatorics and set theory. Understanding how constraints affect function mappings can help in solving more advanced problems in these areas.

We call a function f: A → B almost injective if for every b ∈ B the number of elements a with f(a) = b is at most two. Let M(n) be the number of almost injective monotone functions from [n] to [n]. What is M(5)?

Explore the problem of counting almost injective monotone functions from [5] to [5]. Using combinatorics and the Stars and Bars theorem, we determine that M(5) equals 16. This problem covers advanced mathematical concepts suitable for undergraduate combinatorics students.

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