Pascal's triangle — combinatorial interpretation

WHAT is Pascal's triangle?

Row 0:            1
Row 1:          1   1
Row 2:        1   2   1
Row 3:      1   3   3   1
Row 4:    1   4   6   4   1
Row 5:  1   5   10  10  5   1

Row 4, position 2 is $6 = \binom{4}{2}$ : there are 6 ways to choose 2 items from 4.

WHY does $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ ? (Derive from scratch)

Step 1 — count ordered selections. Choose the 1st item: $n$ ways. 2nd: $n-1$ ways. ... $k$ -th: $n-k+1$ ways. $\text{ordered} = n(n-1)\cdots(n-k+1) = \frac{n!}{(n-k)!}$ Why this step? Each pick removes one option, so the counts drop by 1 each time.

Step 2 — remove the ordering. Any fixed set of $k$ items can be arranged in $k!$ different orders, all of which we counted separately. $\binom{n}{k} = \frac{\text{ordered}}{k!} = \frac{n!}{k!\,(n-k)!}$ Why this step? We wanted sets, not sequences; a set of size $k$ was counted $k!$ times.

WHY is each entry the sum of the two above it? (Pascal's rule)

Combinatorial proof (the honest way). Count subsets of size $k$ from $\{1,2,\dots,n\}$ . Focus on element $n$ :

Subsets that contain $n$ : the other $k-1$ members come from the remaining $n-1$ elements → $\binom{n-1}{k-1}$ ways.
Subsets that don't contain $n$ : all $k$ members come from the remaining $n-1$ elements → $\binom{n-1}{k}$ ways.

These two cases are mutually exclusive and cover every subset, so add them (Addition Principle): $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$

Why this works: every possible selection either uses $n$ or it doesn't — no overlap, no gaps.

Figure — Pascal's triangle — combinatorial interpretation

WHY does each row sum to $2^n$ ?

Combinatorial proof. The left side counts subsets of every size $k=0,1,\dots,n$ — i.e. all subsets of an $n$ -element set. But we can also count all subsets directly: for each of the $n$ elements decide in or out (2 choices), giving $2^n$ . Two counts of the same thing must be equal.

Algebraic cross-check. Put $x=y=1$ in the Binomial Theorem $(x+y)^n=\sum \binom{n}{k}x^{n-k}y^k$ : $2^n=(1+1)^n=\sum_{k=0}^n\binom{n}{k}.$

Worked examples

Steel-manned mistakes

Recall Feynman: explain to a 12-year-old

Imagine you have some toys and want to know how many different handfuls you can grab. Pascal's triangle is a cheat-sheet where each number says "this many handfuls." The cool trick: to fill in a new number, just add the two numbers sitting above it — because any new toy is either in your handful or not, and those two possibilities are exactly the two numbers above. Add up a whole row and you get how many handfuls of every size there are: $2$ times itself once for each toy, because every toy says yes-or-no.

Flashcards

What does the entry in row

n

, position

k

of Pascal's triangle equal?

\binom{n}{k}

, the number of ways to choose

k

items from

n

State the formula for

\binom{n}{k}

\dfrac{n!}{k!\,(n-k)!}

Why divide by

k!

in the combination formula?

Ordered picks over-count each set by its

k!

orderings; we want unordered sets.

State Pascal's rule.

\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}

Give the combinatorial proof of Pascal's rule.

Fix element

n

: subsets containing it

\binom{n-1}{k-1}

+ subsets not containing it

\binom{n-1}{k}

; disjoint and exhaustive.

Why is

\binom{n}{k}=\binom{n}{n-k}

Choosing which

k

to take = choosing which

n-k

to leave.

What is

\sum_{k=0}^n\binom{n}{k}

and why?

2^n

; it counts all subsets (each element in or out).

What is

\sum_{k=0}^n(-1)^k\binom{n}{k}

for

n\ge1

0

; even-sized and odd-sized subsets are equal in number.

Difference between

^nP_k

and

\binom{n}{k}

^nP_k

counts ordered selections;

\binom{n}{k}=\,^nP_k/k!

counts unordered.

Connections

Binomial Theorem — coefficients in $(x+y)^n$ are the triangle's rows.
Combinations and Permutations — $\binom{n}{k}$ vs $^nP_k$ .
Addition and Multiplication Principles — foundation of both proofs.
Factorials — building block of the formula.
Hockey Stick Identity — another diagonal-sum pattern in the triangle.
Fibonacci numbers — shallow diagonals of Pascal's triangle sum to Fibonacci.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Pascal's triangle koi jaadu nahi hai — ye seedha-seedha counting ka table hai. Row $n$ ki position $k$ pe jo number likha hai, wo batata hai ki $n$ cheezon me se $k$ cheezein kitne tareekon se choose kar sakte ho, yaani $\binom{n}{k}$ . Isliye pizza toppings, team selection, subsets — sab yahin se count ho jaate hain.

Formula $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ ka logic simple hai: pehle ordered tarike se pick karo ( $n(n-1)\cdots$ ), phir $k!$ se divide kar do kyunki same set ko $k!$ baar ginn liya (order to matter nahi karta sets me). Yahi sabse badi mistake hoti hai — students $k!$ se divide karna bhool jaate hain, kyunki pick karna one-by-one ordered feel hota hai. Yaad rakho: $\{A,B\}$ aur $\{B,A\}$ ek hi selection hai.

Pascal's rule $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$ ka asli proof bhi counting se hai: last element ko dekho — ya to wo tumhare group me hai (baaki $k-1$ choose karo), ya nahi hai (poore $k$ baaki se choose karo). Ye do cases overlap nahi karte, isliye add kar do. Isi wajah se upar wale do numbers add karke neeche wala aata hai. Aur ek pura row add karo to $2^n$ milta hai, kyunki har element ka bas do choice hai — in ya out. Bas itna samajh lo, to poora triangle khud bana loge, ratne ki zaroorat nahi.

Pascal's triangle — combinatorial interpretation

WHAT is Pascal's triangle?

WHY does $\binom{n}{k}=\frac{n!}{k!(n-k)!}$ ? (Derive from scratch)

WHY is each entry the sum of the two above it? (Pascal's rule)

WHY does each row sum to $2^n$ ?

Worked examples

Steel-manned mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Sequences & Series

Connections