Permutations, Combinations, Probability, Sample Size & P-Value Calculator

Compute nCr, nPr, binomial probability, normal p-values from Z-scores, and 95% confidence sample size.

Combinations C(10,3), Permutations P(10,3) & P-Value
nCr = 120 • nPr = 720
Binomial P(X=3) = 11.72% • Two-Tailed p-value (Z=1.96) = 0.05 • 95% Sample Size = 385
Combinations C(10, 3) 120
Permutations P(10, 3) 720
Two-Tailed p-Value (Z = 1.96) 0.05
Cochran Sample Size (±5%) 385 respondents
Inputs
⚡ Scenarios:
Total population set size or Bernoulli trials.
Subset size selected or number of successes (r ≤ n).
Probability p ∈ (0,1) for binomial & sample-size calculation.
Standard normal Z-score (e.g. 1.96 corresponds to p = 0.05 two-tailed).
%
Desired 95% confidence interval half-width (e.g. ±5%).
Results & Graph
Combinations C(10,3), Permutations P(10,3) & P-Value
nCr = 120 • nPr = 720
Binomial P(X=3) = 11.72% • Two-Tailed p-value (Z=1.96) = 0.05 • 95% Sample Size = 385
Combinations C(10, 3)
120
Permutations P(10, 3)
720
Two-Tailed p-Value (Z = 1.96)
0.05
Significant at α = 0.05
Cochran Sample Size (±5%)
385 respondents

Combinatorics & Sample Size Scale

Combinations (nCr) vs. Permutations (nPr) vs. Cochran Sample Size
Graph 1 • Standard Normal Z-Distribution & Critical Tail Cutoffs Visualizes ±Z test statistic thresholds on N(0,1)
−Z = −1.96 +Z = +1.96 Standard Normal Distribution N(0, 1) • Critical Cutoffs at ±1.96
Graph 2 • Combinatorial Growth Magnitude (Log₁₀ Scale) Compares r!, Combinations nCr, Permutations nPr, and n!
Combinations C(10,3) 120.00 log₁₀ Permutations P(10,3) 720.00 log₁₀ 95% Sample Size 385.00 log₁₀
Graph 3 • Survey Sample Size (n) Required Across Margins of Error n = 1.96² × p(1−p) ÷ e²
0 5.2k 10.4k ±10% MoE: 97 97 ±10% MoE ±7% MoE: 196 196 ±7% MoE ±5% MoE: 385 385 ±5% MoE ±3% MoE: 1,068 1.1k ±3% MoE ±2% MoE: 2,401 2.4k ±2% MoE ±1% MoE: 9,604 9.6k ±1% MoE
Summary: C(10,3) = 120, P(10,3) = 720, and two-tailed p-value for Z=1.96 is 0.05.
Summary: Choosing 3 items from 10 yields 120 unordered combinations (nCr) and 720 ordered permutations (nPr). With trial success probability p = 0.5, the binomial probability of exactly 3 successes in 10 trials is 11.72%. For Z = 1.96, the two-tailed normal p-value is 0.05, and Cochran’s 95% confidence sample size for ±5% margin of error is 385.
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Formula & Mathematical Methodology: Combinatorics, Binomial PMF, Normal CDF & Cochran Sample Size

nCr = n!r!(n − r)!,   nPr = n!(n − r)!,   nsample = 1.962 · p(1 − p)e2
nCr = n! / [r!(n−r)!]; nPr = n! / (n−r)!; P(X=r) = nCr × p^r (1−p)^(n−r); n_sample = (1.96² × p(1−p)) / e²
  • nCr vs nPr: Unordered combinations vs. ordered permutations without replacement
  • e: Margin of error as a decimal (e.g., 0.05 for ±5%)

Rounding: Exact combinatorics; p-values rounded to 4–5 decimal places.

Worked Example: n = 10, r = 3, p = 0.5, Z = 1.96, Margin of Error = ±5%

Compute combinations, permutations, p-value, and required survey sample size.

  1. Permutations P(10,3) = 10 × 9 × 8 = 720; Combinations C(10,3) = 720 ÷ 6 = 120.
  2. Two-tailed p-value at Z = 1.96 is p = 0.0500.
  3. Cochran sample size at 95% confidence (±5% margin, p=0.5) = (3.8416 × 0.25) ÷ 0.0025 = 385 respondents.
Result: nCr = 120, nPr = 720, p-value = 0.0500, and required sample size = 385.

Frequently Asked Questions

Why is 385 the classic survey sample size for a ±5% margin of error?

At 95% confidence (Z = 1.96) and maximum variance (p = 0.5), Cochran’s formula gives 1.96² × 0.25 ÷ 0.05² = 384.16, which rounds up to 385 participants.

Mathematics & Statistics · Probability & Inferential Statistics

Permutations, Combinations, Probability, Sample Size & P-Value Calculator: inputs, method, and how to read the result

Compute nCr, nPr, binomial probability, normal p-values from Z-scores, and 95% confidence sample size.

Choose inputs for the case you are checking

The defaults are illustrative. Replace them with values for your case and keep the units shown beside each field.

  • Total Items / Trials (n): Total population set size or Bernoulli trials.
  • Items Chosen / Successes (r): Subset size selected or number of successes (r ≤ n).
  • Single-Trial Success Probability (p): Probability p ∈ (0,1) for binomial & sample-size calculation.
  • Test Z-Statistic (For P-Value): Standard normal Z-score (e.g. 1.96 corresponds to p = 0.05 two-tailed).

What the calculation includes

With the default values shown, Combinations C(10,3), Permutations P(10,3) & P-Value is nCr = 120 • nPr = 720. The calculation follows Combinatorics, Binomial PMF, Normal CDF & Cochran Sample Size; the formula and variable definitions above show how the inputs combine.

One-input check: With the other defaults held constant, changing Total Items / Trials (n) from 10 to 11 changes Combinations C(10,3), Permutations P(10,3) & P-Value (nCr = 120 • nPr = 720) to Combinations C(11,3), Permutations P(11,3) & P-Value (nCr = 165 • nPr = 990). This is a comparison inside the calculator's stated model, not a prediction.

Where to use caution

  • Sampling without replacement for nCr/nPr; infinite population Cochran formula for sample size.
  • Factorial calculation is bounded at n ≤ 170.

Scope: Global Statistical Standard Check the assumptions and limitations against your situation before relying on the estimate.

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