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 formula requires that all outcomes in the sample space are equally likely — for instance, each card draw or die roll. When outcomes have unequal probabilities (e.g., biased coins, weighted sampling), you must sum probabilities directly instead of counting. With that caveat, this formula is the backbone of nearly every discrete 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 a specific suit, say spades (all 5 cards from the 13 spades), is

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

Since there are 4 suits, the probability of a flush in any suit is $4 \times 0.000495 \approx 0.00198$ (excluding straight flushes).

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.