Combinations — nCr, Pascal's triangle: Visual walkthrough

This is the parent topic seen entirely through pictures. If a word like factorial or permutation is new, we build it here — nothing is assumed.

Step 1 — What "choosing" even means (order vs no-order)

WHAT. We have $n$ distinct objects. Distinct means tell-apart-able — three different candies, five different people. We want to pick a small pile of $r$ of them. The whole question of this page: how many different piles are possible?

WHY the fuss about order. There are two different questions hiding here:

Ordered: "who is 1st, who is 2nd, who is 3rd?" — a line-up.
Unordered: "who is simply IN the pile?" — a pile has no front or back.

A combination answers the unordered one. The symbol we are chasing is

$(r n)$

read aloud as " $n$ choose $r$ ". Here $n$ (top number) is how many we have to pick from, and $r$ (bottom number) is how many we take. That is all it means so far — a count.

PICTURE. Look at the figure: the same three candies form one pile on the right, but six line-ups on the left. Combinations count the single pile; the six line-ups are what we must learn to divide away.

Figure — Combinations — nCr, Pascal's triangle

Step 2 — First count the ORDERED picks (this is $^{n} P_{r}$ )

WHAT. Before we can un-count order, we count with order. Imagine filling $r$ empty slots, left to right, by grabbing objects.

WHY start with order. Ordered picking is easy — it is just a chain of multiplications, one factor per slot. This ordered count is called a permutation, written $^{n} P_{r}$ (see Permutations — nPr).

PICTURE. The figure shows $r$ slots being filled:

$^{n} P_{r} = slot 1: any of n n \times slot 2: one used up (n - 1) \times slot 3 (n - 2) \times \dots \times slot r (n - r + 1)$

Term by term: the first slot has $n$ choices; once one object is placed it is gone, so the second slot has $n - 1$ ; and so on. After filling $r$ slots we have used $r$ objects, so the last factor is $n - r + 1$ (count down $r$ steps starting at $n$ ).

Step 3 — Spot the overcounting

WHAT. Take one fixed pile, say ${A, B, C}$ . How many of our ordered line-ups produced this same pile?

WHY this is the crux. Every unordered pile got counted once for each way its own members can be lined up. So $^{n} P_{r}$ is not the pile-count — it is the pile-count multiplied by the number of internal orderings.

PICTURE. The figure lays the pile ${A, B, C}$ in the centre and its six line-ups fanning out around it: $A B C, A C B, B A C, B C A, C A B, C B A$ . Six spokes, one hub. Every hub (pile) has the same number of spokes.

How many spokes? The number of ways to line up $r$ chosen objects among themselves — that is $^{r} P_{r} = r!$ .

Step 4 — Divide the order away: the master formula

WHAT. If every pile was counted exactly $r!$ times inside $^{n} P_{r}$ , then to recover the true pile-count we simply divide by $r!$ .

WHY division and not subtraction. The overcount is multiplicative: total ordered $=$ (piles) $\times$ ( $r!$ ). Undoing a multiplication is a division. Subtraction would only remove a fixed amount, not a per-pile factor.

PICTURE. The figure shows the hub-and-spoke cloud collapsing: many spokes squeeze into one dot — "un-shuffling" $r!$ line-ups back into a single pile.

$(r n) = \frac{^{n} P _{r}}{r !} = \frac{1}{r !} \cdot \frac{n !}{( n - r )!} = \frac{n !}{r ! ( n - r )!}$

Term by term:

$n!$ — everything we could arrange,
$(n - r)!$ — the tail of objects not picked, divided out because we never chose them,
$r!$ — the internal orderings of the pile, divided out because we don't care about order.

Step 5 — The degenerate cases (never skip these)

WHAT. Test the machine at its edges: $r = 0$ , $r = n$ , and $r = 1$ .

WHY. A formula you trust must give sane answers at the extremes. If it breaks there, it is wrong.

PICTURE. The figure shows three tiny scenes: an empty pile, a "take-everything" pile, and a single-object pile.

$r = 0$ (choose nothing): $(0 n) = \frac{n !}{0 ! n !} = \frac{n !}{1 \cdot n !} = 1$ . There is exactly one empty pile. This needs $0! = 1$ — the reason that convention exists (see Factorials and 0!).
$r = n$ (choose everything): $(n n) = \frac{n !}{n ! 0 !} = 1$ . Exactly one way to take them all.
$r = 1$ : $(1 n) = \frac{n !}{1 ! ( n - 1 )!} = n$ . Picking one object $=$ $n$ choices. Obvious — good, the formula agrees.

Step 6 — The hidden symmetry $(r n) = (n - r n)$

WHAT. Choosing which $r$ to keep is the same act as choosing which $n - r$ to leave behind. One decision, two names.

WHY it must hold. Look at the formula: swapping $r$ and $n - r$ swaps the two denominators $r!$ and $(n - r)!$ — but multiplication does not care about order, so the value is unchanged.

$(r n) = \frac{n !}{r ! ( n - r )!} = \frac{n !}{( n - r )! r !} = (n - r n)$

PICTURE. The figure splits $n$ objects with a dividing line: everything on the left is "kept" ( $r$ ), everything on the right is "tossed" ( $n - r$ ). Sliding the line makes one side the mirror of the other.

Step 7 — Stacking the answers: Pascal's Triangle

WHAT. Put $(r n)$ into a grid — row $n$ , position $r$ — and a triangle of numbers appears.

WHY it self-builds. Fix one special object (call it Ravi). Any pile of size $r$ either contains Ravi or not — two non-overlapping, all-covering cases:

Ravi IN: choose the other $r - 1$ from the remaining $n - 1$ → $(r - 1 n - 1)$ .
Ravi OUT: choose all $r$ from the remaining $n - 1$ → $(r n - 1)$ .

Add the disjoint cases:

$(r n) = (r - 1 n - 1) + (r n - 1)$

So every entry is the sum of the two directly above it. That is why the triangle grows by simple addition — no formula needed once the top rows exist (see Pascal's Triangle patterns and Binomial Theorem).

PICTURE. The figure shows the triangle with two parents $(r - 1 n - 1)$ and $(r n - 1)$ flowing down into their child $(r n)$ , and a Ravi-in / Ravi-out split beside it.

The one-picture summary

WHAT. One figure compresses the whole journey: ordered picks ( $^{n} P_{r}$ ) → collapse each pile's $r!$ line-ups → divide → $(r n)$ → stack into Pascal's Triangle.

Recall Feynman retelling (plain words)

Suppose you can pocket $3$ of $5$ different candies. First pretend order matters and grab them into a line: $5 \times 4 \times 3 = 60$ line-ups. But a pocket doesn't care about order — the same $3$ candies can be lined up $3 \times 2 \times 1 = 6$ ways, and all $6$ describe the same pocket. So I over-counted by $6$ . Divide: $60/6 = 10$ pockets. That "divide-by-the-shuffles" is the combination formula. If I pocket $0$ candies there's exactly $1$ way (the empty pocket) — that's why $0! = 1$ . And Pascal's triangle is just this: to fill any box, pick one special candy; it's either in the pocket or not, so add the two boxes leaning over the box. Order-strip, then case-split — the whole subject in two moves.

Divide, not subtract — why?

The overcount is multiplicative: ordered

=

piles

\times r!

. Undo multiplication by dividing.

Role of

(n - r)!

in the formula?

It cancels the tail of objects never chosen, leaving only the

r

falling factors of

^{n} P_{r}

(2 6) = ?

\frac{6 \cdot 5}{2} = 15

Connections

Permutations — nPr — Step 2 is $^{n} P_{r}$ ; combinations just strip its order.
Factorials and 0! — Step 5 needs $0! = 1$ to make $(0 n) = 1$ .
Binomial Theorem — these triangle entries are the coefficients of $(a + b)^{n}$ .
Pascal's Triangle patterns — Step 7's addition law generates every pattern.
Probability — counting outcomes — piles are the equally-likely unordered outcomes.