Let be the binary linear code
How many binary linear codes are there which are equivalent (under coordinate permutations) to ?
[THOUGHT] I need to determine the number of binary linear codes equivalent to C under coordinate permutations. The code C is defined as all 16-bit vectors where each bit is repeated twice: (a1,a1,a2,a2,...,a8,a8).
Two codes are equivalent if one can be obtained from the other by permuting the coordinates. So, I need to find the number of distinct codes that can be formed by permuting the coordinates of C.
First, I should understand the structure of C. The code has 8 pairs of identical bits. When we permute the coordinates, we are essentially permuting these pairs. However, within each pair, the two bits are identical, so swapping them doesn't change the code.
The problem reduces to counting the number of distinct ways to partition 16 positions into 8 unordered pairs, where the order of the pairs doesn't matter, but the order within each pair doesn't matter either (since the bits are identical).
This is equivalent to counting the number of perfect matchings in a complete graph with 16 vertices, divided by the number of ways to order the pairs (since the pairs are indistinct).
The number of ways to partition 2n objects into n unordered pairs is given by (2n-1)!! = (2n)! / (2^n * n!), where !! is the double factorial.
For n=8, this is (16)! / (2^8 * 8!).
Let me compute this value using Python. [/THOUGHT] [PYTHON] import math n = 8 result = math.factorial(2 * n) // (2 ** n * math.factorial(n)) print(result) [/PYTHON]
[THOUGHT] The Python output confirms that the number of distinct binary linear codes equivalent to C under coordinate permutations is 2027025. This is the number of ways to partition 16 positions into 8 unordered pairs, which corresponds to the number of distinct codes obtainable by permuting the coordinates of C. [/THOUGHT] \boxed{2027025}