Pascal's triangle — combinatorial interpretation: Worked examples

This page is a drill through every kind of counting question Pascal's triangle can throw at you. We start by mapping out all the case-classes in one table, then work an example for each cell. If you have not yet met $(k n)$ (read " $n$ choose $k$ "), the count of ways to choose $k$ things from $n$ things, read the parent note first — here we assume you know it means a count and we push it into every corner.

For the whole page we keep this reference triangle in view:

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 6: 1  6   15  20  15  6   1

The scenario matrix

Every problem below is one cell in this grid. The point of the grid is: you will never meet a Pascal-triangle situation that isn't one of these rows.

Cell	Case class	What makes it tricky	Example
A	Ordinary interior entry	pick a middle number cleanly	Ex 1
B	Edge / degenerate $k = 0$ or $k = n$	"choose nothing" or "choose all" — is that $1$ or $0$ ?	Ex 2
C	Out-of-range $k < 0$ or $k > n$	asks for the impossible	Ex 2
D	Symmetry short-cut $(k n) = (n - k n)$	when $k$ is big, flip it	Ex 3
E	Build a row by Pascal's rule (no factorials)	add the two parents	Ex 4
F	Row sum $= 2^{n}$ (all subsets)	"any number, including none"	Ex 5
G	Alternating sum $= 0$ (even vs odd)	signs flip; must handle $n = 0$ separately	Ex 6
H	Real-world word problem	translate English → $(k n)$	Ex 7
I	Exam twist (Hockey Stick diagonal)	a diagonal sum in disguise	Ex 8
J	Limiting / growth behaviour	which entry is biggest? how fast do rows grow?	Ex 9

Ex 1 — Cell A: an ordinary interior entry

Forecast: guess before reading — is it single digits or does it break into the teens? Write your guess.

Read it off the triangle. Row 6 is $1, 6, 15, 20, 15, 6, 1$ . Counting positions from $k = 0$ , position $2$ is $15$ . Why this step? The triangle is a lookup table of $(k n)$ , so for small $n$ reading beats computing.
Compute from the formula. $(2 6) = \frac{6 !}{2 ! 4 !} = \frac{6 \cdot 5}{2} = \frac{30}{2} = 15$ . Why this step? $(2 6)$ = (number of ordered pairs $6 \cdot 5 = 30$ ) divided by the $2! = 2$ orderings of each pair, because a set ${A, B}$ was counted as both $A B$ and $B A$ .

Verify: the two methods give the same $15$ . Sanity: it must be a whole number (you can't choose "half a way"), and $15 < 30$ (unordered $<$ ordered), both true.

Ex 2 — Cells B & C: degenerate and impossible inputs

Forecast: two of these are $1$ , two are $0$ . Which is which?

$(0 5)$ — choose nothing. There is exactly one way to pick no items: the empty selection. So $(0 5) = 1$ . Why this step? "How many ways to do nothing?" has one answer — do nothing. Formula agrees: $\frac{5 !}{0 ! 5 !} = \frac{1}{1} = 1$ since $0! = 1$ .
$(5 5)$ — choose all. There is exactly one way to take everything: take everything. So $(5 5) = 1$ . Why this step? By symmetry $(5 5) = (0 5) = 1$ — "leave nothing" = "take nothing's mirror".
$(- 1 5)$ — choose $- 1$ items. You cannot select a negative number of things. There are zero such selections: $(- 1 5) = 0$ . Why this step? $k$ must satisfy $0 \leq k \leq n$ ; outside that range the count is defined as $0$ (nothing to count).
$(7 5)$ — choose 7 from 5. You only have $5$ items, so you can't grab $7$ . Zero ways: $(7 5) = 0$ . Why this step? Same range rule from the other side: $k > n$ is impossible.

Verify: $(0 5) = (5 5) = 1$ (the two outer $1$ s bordering every row of the triangle), and out-of-range entries are the invisible $0$ s sitting just outside each row that make Pascal's rule work at the edges.

Ex 3 — Cell D: symmetry short-cut

Forecast: will the answer be huge or small? (Hint: $18$ is close to $20$ .)

Flip using symmetry. $(18 20) = (20 - 18 20) = (2 20)$ . Why this step? Choosing which $18$ to take is the same act as choosing which $2$ to leave — same choices, same count. Flipping turns a long product into a short one.
Compute the easy side. $(2 20) = \frac{20 \cdot 19}{2} = \frac{380}{2} = 190$ . Why this step? Ordered pairs $20 \cdot 19 = 380$ , divided by the $2!$ orderings of each pair.

Verify: $(18 20) = 190$ . Sanity: leaving out $2$ of $20$ — there are clearly $(2 20)$ pairs to leave out, and the symmetry $(k n) = (n - k n)$ guarantees equality.

Ex 4 — Cell E: build a row with Pascal's rule (no factorials)

Forecast: row 7 will have how many entries — $7$ or $8$ ?

Look at the figure: each new entry drops down from the sum of the two entries directly above it.

Figure — Pascal's triangle — combinatorial interpretation

The borders are always $1$ . $(0 7) = (7 7) = 1$ (one way to choose none / all). Why this step? Every row starts and ends with $1$ ; these come from the invisible outer $0 + 1$ .
Add adjacent parents. $(1 7) = 1 + 6 = 7$ ; $(2 7) = 6 + 15 = 21$ ; $(3 7) = 15 + 20 = 35$ ; $(4 7) = 20 + 15 = 35$ ; $(5 7) = 15 + 6 = 21$ ; $(6 7) = 6 + 1 = 7$ . Why this step? Pascal's rule $(k n) = (k - 1 n - 1) + (k n - 1)$ : a size- $k$ subset of $7$ either uses element $7$ (choose $k - 1$ more from $6$ ) or doesn't (choose $k$ from $6$ ). Disjoint + exhaustive → add.
Assemble. Row 7 $= 1, 7, 21, 35, 35, 21, 7, 1$ .

Verify: the row is a palindrome ( $1, 7, 21, 35, 35, 21, 7, 1$ ) as symmetry demands, and it has $7 + 1 = 8$ entries. Cross-check one by formula: $(3 7) = \frac{7 \cdot 6 \cdot 5}{3 !} = \frac{210}{6} = 35$ . ✓

Ex 5 — Cell F: row sum $= 2^{n}$ (all subsets)

Forecast: guess an order of magnitude — dozens? hundreds?

Translate to subsets. A sandwich is a choice of which fillings are in. Each of the $6$ fillings is independently in or out. Why this step? "Any number of fillings, including none" is exactly "any subset of the $6$ fillings".
Count by row sum. $k = 0 \sum 6 (k 6) = 2^{6} = 64$ . Why this step? Summing $(k 6)$ over all sizes $k$ counts subsets of every size; equivalently each filling says yes/no ( $2$ choices), giving $2^{6}$ . Two counts of the same thing must match.

Verify: add row 6 directly: $1 + 6 + 15 + 20 + 15 + 6 + 1 = 64$ . ✓ And $2^{6} = 64$ . The "plain" sandwich is the $k = 0$ term (the single $1$ ), so it is counted.

Ex 6 — Cell G: alternating sum $= 0$ (even vs odd), and the $n = 0$ trap

Forecast: part (b) is the trap — is it also $0$ ?

Write the alternating sum for $n = 5$ . $(0 5) - (1 5) + (2 5) - (3 5) + (4 5) - (5 5) = 1 - 5 + 10 - 10 + 5 - 1$ . Why this step? The $(- 1)^{k}$ flips sign every step, tagging even-sized subsets $+$ and odd-sized subsets $-$ .
Add. $1 - 5 + 10 - 10 + 5 - 1 = 0$ . Why this step? The pairing $(1, - 1), (- 5, 5), (10, - 10)$ cancels — this says #(even subsets) $=$ #(odd subsets) $= 16$ each, and $16 + 16 = 32 = 2^{5}$ .
Now $n = 0$ . The only term is $k = 0$ : $(- 1)^{0} (0 0) = 1$ . The sum is $1$ , not $0$ . Why this step? The identity $\sum_{k} (- 1)^{k} (k n) = 0$ holds only for $n \geq 1$ . At $n = 0$ the empty set is the only subset and it is even-sized — there is no odd subset to cancel it. Algebra agrees: $(1 - 1)^{0} = 0^{0} = 1$ .

Verify: $1 - 5 + 10 - 10 + 5 - 1 = 0$ (part a), and for $n = 0$ the value is $1$ (part b). The two halves $16 = 16$ split row 5's $32$ evenly, confirming even = odd for $n \geq 1$ .

Ex 7 — Cell H: a real-world word problem

Forecast: you must choose women and men — do you add or multiply the two counts?

Split the job into independent choices. Choose $2$ women from $5$ : $(2 5)$ ways. Choose $2$ men from $4$ : $(2 4)$ ways. Why this step? "Exactly $2$ women" fixes the split: $2$ women $+ 2$ men $= 4$ members. The two selections don't interfere.
Compute each. $(2 5) = 10$ (row 5, position 2). $(2 4) = 6$ (row 4, position 2). Why this step? Read straight off the triangle.
Multiply (Multiplication Principle). $10 \times 6 = 60$ committees. Why this step? For each of the $10$ women-choices you may pair any of the $6$ men-choices — independent stages multiply, they do not add.

Verify: $(2 5) \cdot (2 4) = 10 \cdot 6 = 60$ . Sanity: the total committees of any composition would be $(4 9) = 126$ ; our "exactly 2 women" slice ( $60$ ) is under that, as it must be. See Addition and Multiplication Principles for why we multiplied here (independent stages) rather than added (mutually exclusive cases).

Ex 8 — Cell I: the exam twist (Hockey Stick diagonal)

Forecast: guess whether the shortcut answer sits outside the entries you added.

Follow the coloured "hockey stick" in the figure: a straight diagonal of entries plus one turn at the tip.

Name the parts. The sum is $3 + 6 + 10 + 15 + 21$ . Why this step? These are $(2 2), (2 3), (2 4), (2 5), (2 6)$ read down the " $2$ "-diagonal.
Apply the Hockey Stick Identity. $i = r \sum n (r i) = (r + 1 n + 1)$ , here $r = 2$ , $n = 6$ : the sum equals $(3 7)$ . Why this step? Repeatedly using Pascal's rule collapses the diagonal to the entry one row below and one step in — the "tip" of the hockey stick. See Hockey Stick Identity.
Read the tip. $(3 7) = 35$ (from row 7 we built in Ex 4).

Verify: direct addition $3 + 6 + 10 + 15 + 21 = 55$ … wait — recount: $3 + 6 = 9$ , $+ 10 = 19$ , $+ 15 = 34$ , $+ 21 = 55$ . And $(3 7) = 35$ ? These disagree, so recheck the identity's index. The diagonal $(2 2), \dots, (2 6)$ sums to $(3 7) = 35$ only if the terms are $(2 2) = 1, (2 3) = 3, (2 4) = 6, (2 5) = 10, (2 6) = 15$ giving $1 + 3 + 6 + 10 + 15 = 35$ . ✓ (Our step 1 mislabelled the first term — $(2 2) = 1$ , not $3$ .) Corrected sum $= 35 = (3 7)$ .

Why this correction matters: $(2 2) = 1$ (choose all $2$ of $2$ ), a degenerate/edge value from Cell B sneaking into a Cell I problem — always start the diagonal at the true top entry.

Ex 9 — Cell J: limiting / growth behaviour

Forecast: is the biggest entry in the middle or near the edge? Does the total double or square?

Find the peak. Row 6 $= 1, 6, 15, 20, 15, 6, 1$ . The largest is the middle entry $(3 6) = 20$ . Why this step? Entries rise to the centre then fall (by symmetry $(k 6) = (6 - k 6)$ ). The central $k = n /2$ term is the mode of the row.
Compare row totals. Row 6 sum $= 2^{6} = 64$ ; row 5 sum $= 2^{5} = 32$ . Why this step? Each new row adds one more element to be "in or out", doubling the number of subsets: $2^{n + 1} = 2 \cdot 2^{n}$ .
State the growth law. Rows grow by a factor of exactly $2$ each step, and the peak entry grows too (from $(3 5) = 10$ to $(3 6) = 20$ — also doubling here). Why this step? This is the seed of why binomial distributions concentrate near the middle for large $n$ .

Verify: peak $= 20 = (3 6)$ ; totals $64 = 2 \cdot 32$ , so the row-total ratio is exactly $2$ . ✓

Recall Which cell was which?

Ordinary entry ::: Ex 1 (Cell A) "Choose none / all / impossible" ::: Ex 2 (Cells B, C) Flip $k$ when it's large ::: Ex 3 (Cell D) Build a row by adding parents ::: Ex 4 (Cell E) "Any number, including none" = $2^{n}$ ::: Ex 5 (Cell F) Even vs odd subsets, and the $n = 0$ exception ::: Ex 6 (Cell G) Word problem: multiply independent stages ::: Ex 7 (Cell H) Diagonal sum = one entry below ::: Ex 8 (Cell I) Biggest entry is the middle; rows double ::: Ex 9 (Cell J)

Connections

Pascal's Triangle — Combinatorial Interpretation — the parent this page drills.
Combinations and Permutations — every $(k n)$ here is an unordered count.
Addition and Multiplication Principles — Ex 7's "multiply, don't add".
Factorials — the $k!$ and $0! = 1$ that fix the degenerate cases (Ex 2).
Binomial Theorem — row sums (Ex 5) and alternating sums (Ex 6) are $x = y = 1$ and $x = 1, y = - 1$ .
Hockey Stick Identity — the diagonal collapse in Ex 8.
Fibonacci numbers — shallow diagonals give another growth pattern beyond Ex 9.