Pascal's triangle

Pascal's Triangle is an infinite triangular array where each number is the sum of the two numbers directly above it:

$111413126131411$

The numbers in Pascal's Triangle are precisely the binomial coefficients $(k n)$ . In row $n$ (starting at row 0), position $k$ (starting at $k = 0$ ), the number represents $(k n)$ .

$(0 4) (0 3) (0 2) (1 4) (0 1) (1 3) (0 0) (1 2) (2 4) (1 1) (2 3) (2 2) (3 4) (3 3) (4 4)$
This connection arises naturally from the recursive definition of binomial coefficients: ¹
$(k n) = (k - 1 n - 1) + (k n - 1)$
Which is exactly how Pascal’s Triangle is constructed: Each number is the sum of the two numbers above it.
The recursive formula represents that to choose $k$ items from $n$ items, we can either:

Take a specific item and choose $k - 1$ more from the remaining $n - 1$ items ( $(k - 1 n - 1)$ )

Leave out that specific item and choose all $k$ from the remaining $n - 1$ items ( $(k n - 1)$ )

aka Pascal’s identity.

Reading binomial coefficients from the triangle:

$(2 4) = 6$ can be found in row 4, position 2, telling us there are 6 ways to choose 2 items from 4 items.
$(1 + x)^{4} = (0 4) x^{0} + (1 4) x^{1} + (2 4) x^{2} + (3 4) x^{3} + (4 4) x^{4} = 1 + 4 x + 6 x^{2} + 4 x^{3} + x^{4}$
These coefficients appear directly in row 4 of Pascal’s Triangle: 1, 4, 6, 4, 1

The sum of all binomial coefficients for fixed $n$ (row of Pascal’s triangle) equals $2^{n}$

$k = 0 \sum n (k n) = 2^{n}$
This makes intuitive sense when thinking about sets: $(k n)$ counts the ways to choose $k$ items from $n$ items, and summing over all possible $k$ gives us the total number of possible subsets of an $n$ -element set, which is $2^{n}$ . Each element either is or isn’t in the subset, giving us two choices for each of the $n$ elements → power set.

In the first few rows:

Row 0: $(0 0) = 1 = 2^{0}$ , i.e. we can choose none of the 0 items.
Row 1: $(0 1) + (1 1) = 1 + 1 = 2 = 2^{1}$ , i.e we can choose all or none of the 1 item.
Row 2: $(0 2) + (1 2) + (2 2) = 1 + 2 + 1 = 4 = 2^{2}$ , i.e. we can choose none, one in two ways, or both items.
Row 3: $(0 3) + (1 3) + (2 3) + (3 3) = 1 + 3 + 3 + 1 = 8 = 2^{3}$ , i.e. we can choose none, have three choices for picking one item, three choices for picking two items, or all three items.

Pascals triangle (the binomial distribution) converges to the normal distribution for large $n$ .

$X \sim binomial (n, p) \to X \sim normal (n p, n pq) \to X \sim poisson (λ) if n pq is not too small if n p or n q is small$
where $p$ is the probability of success, $q = 1 - p$ is the probability of failure, and $n$ is the number of trials.
This is true $n pq$ is not too small, i.e. success and failure are not too rare.
If they are, the binomial distribution converges to the poisson distribution, where $λ = n p$ is the average number of successes.

E.g the value $(2 3) = 3$ comes from adding the two numbers above it: $(2 3) = (1 2) + (2 2) = 2 + 1 = 3$ ↩

Max Wolf's Second Brain

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Pascal's triangle

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Max Wolf's Second Brain

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Pascal's triangle

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