TL;DR

The number of ordered arrangements of r items from n distinct items: \( P(n,r) = \frac{n!}{(n-r)!} \). For all n items, \( P(n,n) = n! \).

By Valenke Exam Prep Team·Last updated 2026-06-03

Permutations

The number of ordered arrangements of r items from n distinct items: \( P(n,r) = \frac{n!}{(n-r)!} \). For all n items, \( P(n,n) = n! \).

Why it matters for interviews

Basic counting is tested in every quant interview. Permutations arise in card problems, seating arrangements, and combinatorial probability questions that require ordered counting.

Definition and Mathematical Foundation

The number of ordered arrangements of r items from n distinct items: \( P(n,r) = \frac{n!}{(n-r)!} \). For all n items, \( P(n,n) = n! \).

Application in Quantitative Finance

Basic counting is tested in every quant interview. Permutations arise in card problems, seating arrangements, and combinatorial probability questions that require ordered counting.

Related Terms

Combinations

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Frequently Asked Questions

When do you use permutations vs combinations?▼

Use permutations when order matters (arrangements, rankings, sequences). Use combinations when order does not matter (choosing subsets, committees, hands of cards).

What about permutations with repetition?▼

If items repeat, the count is \( \frac{n!}{n_1! n_2! \cdots n_k!} \), the multinomial coefficient. For example, arrangements of MISSISSIPPI: \( 11!/(1!4!4!2!) \).

How do permutations connect to probability?▼

Many probability problems reduce to counting favorable permutations over total permutations. For example, the probability of a specific poker hand is the number of favorable arrangements divided by total arrangements.