Binomial Coefficient

Context: FIT1058_MOC · counts the size- $k$ subsets of an $n$ -set · the size-layers of the Power Set · obeys Pascal’s identity

Quick Revision

🎯 Objective: $(k n)$ = number of $k$ -subsets of an $n$ -set ➔ “n choose k”.

📦 Core Components: closed form $\frac{n !}{( n - k )! k !}$ ➔ symmetry ➔ Pascal’s identity.

⚡ Key Constraint: $\sum_{k} (k n) = 2^{n}$ (size-layers of the Power Set).

📝 Core

1. The Coefficient

Definition ➔ $(k n)$ = number of $k$ -element subsets of an $n$ -set.
Sum ➔ $\sum_{k = 0}^{n} (k n) = 2^{n}$ (Power Set).

2. Closed Form

Ordered first ➔ $n (n - 1) \dots (n - k + 1) = \frac{n !}{( n - k )!}$ .
Correct for order ➔ each subset counted $k!$ times ⟹ $(k n) = \frac{n !}{( n - k )! k !}$ .

3. Identities

Symmetry ➔ $(k n) = (n - k n)$ (include vs exclude).
Pascal ➔ $(k n) = (k - 1 n - 1) + (k n - 1)$ (contains vs omits a fixed element).

Key identities:

$(k n) = \frac{n ( n - 1 ) \dots ( n - k + 1 )}{k !} = \frac{n !}{( n - k )! k !}$ $(k n) = (n - k n), (k n) = (k - 1 n - 1) + (k n - 1)$

⚖️ Core Decision Matrix

Case	Value	Reason
$(0 n)$	1	empty subset
$(n n)$	1	whole set
$(1 n)$	$n$	single elements
$(n - 1 n)$	$n$	leave one out

When It Flips: two computation routes — closed form (direct) or Pascal's identity (recursive, no large factorials, good for many coefficients). Outside $0 \leq k \leq n$ , $(k n) = 0$ .

📊 Exam Execution Trace

Manual Execution Trace

Pascal’s triangle rows:

Step / State	$n$	Row
0 (Init)	0	1
1	1	1 1
2	2	1 2 1
3	3	1 3 3 1

⚠️ Common Mistakes

💡 Divide by $k!$ ➔ ordered selection $\frac{n !}{( n - k )!}$ counts each subset $k!$ times; forgetting the division confuses permutations with combinations.

🧠 Active Recall

Derive $(k n) = \frac{n !}{( n - k )! k !}$ from first principles.

Hint: Order then correct.

Answer

Short answer: Ordered $k$ -selections number $\frac{n !}{( n - k )!}$ ; each subset arises from $k!$ orderings, so divide by $k!$ .

Why: Overcount ➔ combinations = permutations ÷ $k!$ .

State and justify Pascal's identity combinatorially.

Hint: Contains vs omits an element.

Answer

Short answer: $(k n) = (k - 1 n - 1) + (k n - 1)$ : subsets with $b$ ( $(k - 1 n - 1)$ ) + subsets without $b$ ( $(k n - 1)$ ).

Why: Disjoint + exhaustive ➔ the two cases add; the rule builds Pascal’s triangle.

Why does $(k n) = (n - k n)$ , and what are $(0 n)$ and $(n n)$ ?

Hint: Include ⟺ exclude.

Answer

Short answer: Choosing $k$ to include = choosing $n - k$ to exclude; $(0 n) = (n n) = 1$ .

Why: Bijection ➔ one way to choose nothing/everything.

🎓 CS × DS Vault

Explorer

Binomial Coefficient

Binomial Coefficient

📝 Core

1. The Coefficient

2. Closed Form

3. Identities

⚖️ Core Decision Matrix

📊 Exam Execution Trace

Manual Execution Trace

⚠️ Common Mistakes

🧠 Active Recall

Graph View

Table of Contents

Backlinks