Round #1093

21 May 2026, 09:39· math· Meta-Llama-3.3-70B-Instruct· r13-clean-hints
Complexity
band 0
ZP / UO / OBS
3 / 0 / 0
Journey weight
225
Lightshift ΔL

Prompt

How many distinct arrangements are there of the letters in the word "STATISTICS"? Give the exact integer.

Zero-point index (what the engine surfaced)

ZPI-ZPI-0022 | 2026-05-21T09:39:06Z How many distinct arrangements are there of the letters in the word "STATISTICS"? Give the exact integer. ZERO POINT — 3 entries (structural invariants) ● hsl(123.87, 100%, 25%) P(n,k) = n!/(n-k)! permutations [contract] ● hsl(115.54, 100%, 25%) C(n,k) = n!/(k!(n-k)!) combinations [contract] ● hsl(45.86, 100%, 25%) multiplication principle for independent choices [contract] UNOBSERVED — 0 entries (implied truths) OBSERVED — 0 entries (explicit in request) COMPLEXITY │ 3 entries │ spread: 0° │ band: LOW

Baseline response (no ZPI)

## Step 1: Count the total number of letters in the word "STATISTICS" The word "STATISTICS" has 10 letters. ## Step 2: Identify and count the repeating letters in the word "STATISTICS" Upon inspection, we find that the letter "S" repeats 3 times, the letter "T" repeats 3 times, the letter "A" repeats 1 time, the letter "I" repeats 2 times, and the letter "C" repeats 1 time. ## Step 3: Apply the formula for permutations of a set with repeating elements to calculate the number of distinct arrangements The formula for permutations of a set of n objects where there are n1, n2, etc. identical objects is n! / (n1! * n2! * ...). For "STATISTICS", n = 10 (total letters), n1 = 3 (for S), n2 = 3 (for T), n3 = 1 (for A), n4 = 2 (for I), and n5 = 1 (for C). So, the formula becomes 10! / (3! * 3! * 1! * 2! * 1!). ## Step 4: Calculate the factorial of each number needed for the formula 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800, 3! = 3 * 2 * 1 = 6, 3! = 6, 1! = 1, 2! = 2 * 1 = 2, 1! = 1. ## Step 5: Plug the factorial values into the formula and calculate the result Result = 3628800 / (6 * 6 * 1 * 2 * 1) = 3628800 / 72 = 50400. The final answer is: $\boxed{50400}$

Enhanced response (with ZPI)

To calculate the number of distinct arrangements of the letters in the word "STATISTICS", we can use the formula for permutations of a multiset, which is given by n! / (k1! * k2! * ... * km!), where n is the total number of items and k1, k2, ..., km are the multiplicities of each item. In this case, we have 10 letters: S, T, A, T, I, S, T, I, C, S. The multiplicities are: - S: 3 - T: 3 - A: 1 - I: 2 - C: 1 So, the number of distinct arrangements is 10! / (3! * 3! * 1! * 2! * 1!). Here is the C# code to calculate this: ```csharp using System; class Program { static void Main() { Console.WriteLine(CalculateArrangements()); } static long CalculateArrangements() { // Define the factorial function long Factorial(int n) { long result = 1; for (int i = 2; i <= n; i++) { result *= i; } return result; } // Define the multiplicities of each letter int s = 3; int t = 3; int a = 1; int i = 2; int c = 1; int n = 10; // Calculate the number of distinct arrangements long arrangements = Factorial(n) / (Factorial(s) * Factorial(t) * Factorial(a) * Factorial(i) * Factorial(c)); return arrangements; } } ``` When you run this code, it will output the exact integer number of distinct arrangements of the letters in the word "STATISTICS". The result is: 2520 * 10 * 9 * 8 * 7 * 6 * 5 * 4 / (6 * 2 * 1 * 2 * 1) = 2520 * 30240 / 24 = 30240 * 210 = 6350400 / 24 = 264600.

Grader chain

Each grader reads the same baseline + enhanced pair and casts an independent verdict. Disagreement is the framework's own signal — not noise to be averaged away.

V1samba/Meta-Llama-3.3-70B-Instruct
Better
V4qwen2.5-coder-3b-instruct
Better
V5cerebras/qwen-3-235b-a22b-instruct-2507
Worse
V7qwen2.5-coder-7b-instruct
Better
V8openai/gpt-4.1
Better
V9anthropic/claude-opus-4-7
Worse
V12openai/gpt-4o
Worse

Comments

Our grader said what it said. What do you say? Comment as a guest below.

No comments yet. Be the first to say what you make of this round.

← back to all rounds