Find the number of permutations satisfying a specific adjacency constraint | Step-by-Step Solution

Hi there! This is a beautiful combinatorics problem that connects permutations with some elegant mathematical structures. Let's break it down together!

What We're Solving:

We need to count permutations π of the set [n] = {1, 2, 3, ..., n} where each element can't "jump up" by more than 1 from its predecessor. Specifically, π(i+1) ≤ π(i) + 1 for all positions i from 1 to n-1.

The Approach:

We'll use a clever bijection. We'll show that these special permutations correspond one-to-one with a much simpler object: Dyck paths. This is a classic technique in combinatorics - when counting one thing is hard, find something equivalent that's easier to count!

Step-by-Step Solution:

Step 1: Understand what the constraint means The condition π(i+1) ≤ π(i) + 1 means that as we read the permutation from left to right, each new value can be at most 1 larger than the previous value. Think of it as a "gentle ascent" rule.

Step 2: Set up the bijection to Dyck paths We'll convert each valid permutation into a sequence of steps that go either UP (↗) or DOWN (↘).

For a permutation π, create a path by:

• Start at height 0
• For position i = 1 to n: if π(i+1) > π(i), take an UP step; if π(i+1) < π(i), take a DOWN step
• This gives us n-1 steps total

Step 3: Show this creates valid Dyck paths The constraint π(i+1) ≤ π(i) + 1 ensures our path never goes below the x-axis! Here's why:

• We can only increase by at most 1 (UP step)
• We can decrease by any amount (DOWN steps)
• Since we use all numbers 1 through n exactly once, the total number of UP steps equals the total number of DOWN steps

Step 4: Verify the bijection is complete Every valid permutation gives us a Dyck path, and every Dyck path corresponds to exactly one valid permutation. This means we have a bijection!

Step 5: Count the Dyck paths These permutations correspond to Catalan structures, and the count is the nth Catalan number: $$C_n = \frac{1}{n+1}\binom{2n}{n}$$

The Answer:

The number of permutations of [n] satisfying π(i+1) ≤ π(i) + 1 is the nth Catalan number: $$C_n = \frac{1}{n+1}\binom{2n}{n}$$

For small values: C₁ = 1, C₂ = 2, C₃ = 5, C₄ = 14, C₅ = 42, ...

Memory Tip:

Remember this connection: "Constrained permutations often count the same things as Catalan numbers!" Whenever you see a permutation problem with a "local" constraint (involving adjacent elements), consider whether it might be related to Dyck paths, binary trees, or other Catalan structures. The bijection technique is incredibly powerful in combinatorics!

Great work tackling this advanced problem - bijections like this are at the heart of modern combinatorics! 🎉