Basic Counting

Why Counting Matters

Every probability calculation rests on one fundamental question: how many outcomes are there? If we can count the total number of equally likely outcomes and the number of favorable ones, then

$P(\text{event}) = \frac{\text{favorable outcomes}}{\text{total outcomes}}.$

This deceptively simple formula is the backbone of nearly every probability interview question you will encounter.

The Multiplication Principle

If one task can be completed in $m$ ways and a second, independent task can be completed in $n$ ways, the two tasks together can be completed in $m \times n$ ways.

Example. A standard die has 6 faces. Rolling two dice produces $6 \times 6 = 36$ equally likely outcomes. The probability of rolling a total of 7 is $6/36 = 1/6$ because exactly six pairs sum to 7: $(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$.

Permutations

A permutation counts ordered arrangements. The number of ways to arrange $k$ items chosen from $n$ distinct items is

$P(n,k) = \frac{n!}{(n-k)!}.$

When $k = n$, this reduces to $n!$, the total number of ways to arrange all $n$ items.

Example. How many ways can 3 traders be assigned to 3 distinct desks? Answer: $3! = 6$.

Combinations

When order does not matter we use combinations. The number of ways to choose $k$ items from $n$ is

$\binom{n}{k} = \frac{n!}{k!\,(n-k)!}.$

This is read "$n$ choose $k$." The $k!$ in the denominator removes the duplicate orderings we counted in the permutation formula.

Example — Poker hands. A standard deck has 52 cards. The number of distinct 5-card hands is

$\binom{52}{5} = \frac{52!}{5!\,47!} = 2{,}598{,}960.$

The probability of being dealt a flush in spades (all 5 cards from 13 spades) is

$\frac{\binom{13}{5}}{\binom{52}{5}} = \frac{1287}{2{,}598{,}960} \approx 0.000495.$

Worked Interview Problem

You have a bag with 5 red and 4 blue balls. You draw 3 at random without replacement. What is the probability that exactly 2 are red?

Step 1 — Total ways to draw 3 from 9: $\displaystyle \binom{9}{3} = 84$.

Step 2 — Favorable: choose 2 red from 5 and 1 blue from 4:

$\binom{5}{2}\binom{4}{1} = 10 \times 4 = 40.$

Step 3 — Probability: $40/84 = 10/21 \approx 0.476$.

Interview Tip: Always state your counting model before computing. Interviewers want to hear "I'm choosing an unordered subset of size $k$ from $n$, so I use $\displaystyle \binom{n}{k}$." Verbalizing this avoids the most common mistake — confusing permutations with combinations.

Stars and Bars (Bonus Tool)

The number of ways to distribute $n$ identical items into $k$ distinct bins is

$\binom{n+k-1}{k-1}.$

This appears frequently when counting portfolio allocations or partitioning trades across accounts.

Key Insight

Counting is the gateway to probability. Master $\displaystyle \binom{n}{k}$ and the multiplication principle and you will be able to set up the vast majority of discrete probability problems you see in interviews.