Count policy-valid ordered character strings across a length range, with optional required disjoint character groups.
How to use this tool
- Choose the event or model and review the assumptions below.
- Enter the labeled inputs.
- Select Calculate and read the labeled outputs.
- Change an input and calculate again; Reset restores the example inputs.
Calculation method
Without group requirements, count=Σ Cᴸ over the allowed lengths. With all groups required, inclusion–exclusion removes strings missing one or more groups.
Assumptions and limits
Counts policy-valid strings with repetition, not the strength of a human-chosen password. Group character sets must be disjoint. No real password is requested, stored or tested. Lengths up to 200, at most 8 groups and 1,000 characters in total.
Worked example
Groups of size 2 and 3 at length 3 allow 125 unrestricted strings. Requiring both groups leaves 90. No actual password is needed for this policy-space calculation.
Worked reading and practical questions
- A Password Policy Count Is Not Automatically Password Entropy
- Count Required Character Groups with Inclusion–Exclusion
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Frequently asked questions
What does this calculator assume?
Counts policy-valid strings with repetition, not the strength of a human-chosen password. Group character sets must be disjoint. No real password is requested, stored or tested. Lengths up to 200, at most 8 groups and 1,000 characters in total.
How do I enter and update the inputs?
Use the labeled fields, then select Calculate. Percent fields use 0 to 100. After editing, calculate again; Reset restores the example inputs.