Combinations — nCr, Pascal's triangle: Worked examples

This page is the workout room for combinations. The parent note built the machinery; here we drag it through every kind of situation a problem can hand you — tiny edge cases, degenerate zeros, symmetry shortcuts, "at least one" complements, real-world word problems, and an exam-style trap.

Before we count anything, one reminder in plain words: $(r n)$ (say " $n$ choose $r$ ") is how many different groups of $r$ things you can pull out of $n$ things when the order you grabbed them in does not matter. Every symbol below is that same idea in disguise.

The scenario matrix

Think of a topic as a machine with dials. Each dial setting is a different kind of problem. If we test every setting, nothing on the exam can surprise us. Here are the settings for combinations:

#	Case class	What makes it tricky	Hit by example
A	Ordinary $0 < r < n$	plain counting, expect a medium number	Ex 1
B	Edge: $r = 0$ (choose nothing)	tempting to say $0$ , truth is $1$	Ex 2
C	Edge: $r = n$ (choose everything)	only one full group exists	Ex 2
D	Degenerate: $r > n$ (asking too much)	impossible — answer must be $0$	Ex 2
E	Symmetry shortcut $r$ close to $n$	huge-looking, actually tiny via $(r n) = (n - r n)$	Ex 3
F	Pascal check — build a value two ways	verifies the addition law	Ex 4
G	"At least one" via complement	many sub-cases collapse to Total − None	Ex 5
H	Split selection (choose from two groups)	multiply independent choices	Ex 6
I	Real-world word problem	translate English → $n$ and $r$	Ex 7
J	Exam twist: permutation-vs-combination trap	order secretly matters (or doesn't)	Ex 8

We now walk through examples 1–8 so that every row above gets covered.

Example 1 — Cell A: the plain vanilla count

Worked example Choose a 4-player team from 9 players

How many different 4-player teams can be formed from 9 distinct players?

Forecast: guess before reading — will it be more or fewer than $100$ ? Jot a number.

Step 1. Recognise: a team has no internal order — swapping two chosen players gives the same team. Why this step? Combinations, not permutations, exactly because order is irrelevant. If we cared who is "captain first," it would be a permutation.

Step 2. Write the falling-factor form (top $r = 4$ factors of $9$ , over $4!$ ): $(4 9) = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 !} = \frac{9 \cdot 8 \cdot 7 \cdot 6}{24} .$ Why this step? The numerator $9 \cdot 8 \cdot 7 \cdot 6$ is the ordered count $^{9} P_{4}$ ; dividing by $4! = 24$ erases the ordering we don't want (the "un-shuffle").

Step 3. Simplify: $\frac{3024}{24} = 126.$

Verify: Cross-check with the symmetry $(4 9) = (5 9)$ ; and against Pascal's row $n = 9$ : $1, 9, 36, 84, 126, 126, 84, 36, 9, 1$ — the fifth entry is indeed $126$ . ✓

Example 2 — Cells B, C, D: the three edge/degenerate cases at once

Worked example

$(0 6)$ , $(6 6)$ , and $(8 6)$ Evaluate all three.

Forecast: which of these do you think equals $0$ ?

Step 1 (Cell B, $r = 0$ ). "How many ways to choose nothing from 6?" There is exactly one way: take the empty set. $(0 6) = \frac{6 !}{0 ! 6 !} = \frac{6 !}{1 \cdot 6 !} = 1.$ Why this step? The value hinges on $0!=1$. If you wrongly thought $0! = 0$ you'd divide by zero — nonsense. There's genuinely one empty group, so $1$ is right.

Step 2 (Cell C, $r = n$ ). "Choose all 6 from 6." Only one full group exists. $(6 6) = \frac{6 !}{6 ! 0 !} = 1.$ Why this step? Same $0! = 1$ engine — and it matches Step 1 by symmetry $(0 6) = (6 6)$ .

Step 3 (Cell D, $r > n$ ). "Choose 8 from 6" is impossible — you can't pull more items than exist. $(8 6) = 0.$ Why this step? In the formula $\frac{6 !}{8 ! ( 6 - 8 )!}$ the term $(6 - 8)! = (- 2)!$ is undefined, which is nature's way of screaming "impossible." The count of impossible things is $0$ .

Verify: The whole $n = 6$ Pascal row is $1, 6, 15, 20, 15, 6, 1$ — its ends are $1$ (matching Steps 1–2), and there is no 9th entry (matching Step 3's $0$ ). ✓

Example 3 — Cell E: symmetry turns a giant into an ant

Example 4 — Cell F: build one value two ways (Pascal's rule live)

Worked example Get

$(3 6)$ from its two Pascal parents — with a picture Forecast: which two numbers in row $n = 5$ do you think add to give it?

Step 1. Compute directly first: $(3 6) = \frac{6 \cdot 5 \cdot 4}{3 !} = \frac{120}{6} = 20.$ Why this step? We want a target to confirm the addition law reproduces.

Step 2. Apply Pascal's rule $(r n) = (r - 1 n - 1) + (r n - 1)$ with $n = 6, r = 3$ : $(3 6) = (2 5) + (3 5) .$ Why this step? Story-proof (from the parent): fix one special person "Ravi." Either Ravi is IN the committee — then pick $r - 1 = 2$ more from the other $5$ , giving $(2 5)$ — or Ravi is OUT — pick all $3$ from the other $5$ , giving $(3 5)$ . These two cases don't overlap and cover everything, so we add. The red cell in the figure is the sum of the two black cells leaning over it.

Step 3. Evaluate the parents: $(2 5) = 10$ and $(3 5) = 10$ , so $10 + 10 = 20.$

Verify: $20 = 20$ ✓ — the direct value equals the Pascal sum. This is exactly the addition law that grows the whole triangle.

Example 5 — Cell G: "at least one" via the complement

Worked example From 5 physicists and 3 chemists, choose 4 scientists with

at least one chemist Forecast: would you rather add up "1 chemist" + "2 chemists" + "3 chemists" separately, or do one subtraction?

Step 1. Count all groups of 4 from the $5 + 3 = 8$ scientists: $(4 8) = \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 !} = \frac{1680}{24} = 70.$ Why this step? "At least one chemist" is messy (many cases), but its opposite — "no chemist at all" — is a single clean count. Start with the whole universe.

Step 2. Count the complement: groups with zero chemists, i.e. all 4 chosen from the 5 physicists: $(4 5) = (1 5) = 5.$ Why this step? Used symmetry $(4 5) = (1 5)$ to make it instant. This is the "none" case we will remove.

Step 3. Subtract: $70 - 5 = 65.$ Why this step? Total groups minus the ones we don't want (no chemist) leaves exactly the "at least one chemist" groups.

Verify (the long way, must agree): count by exact number of chemists:

1 chemist: $(1 3) (3 5) = 3 \cdot 10 = 30$
2 chemists: $(2 3) (2 5) = 3 \cdot 10 = 30$
3 chemists: $(3 3) (1 5) = 1 \cdot 5 = 5$

Sum $= 30 + 30 + 5 = 65.$ ✓ Matches the complement method.

Example 6 — Cell H: split selection (multiply independent choices)

Worked example A committee of 2 boys AND 3 girls from 5 boys and 6 girls

Forecast: do we add the two counts or multiply them?

Step 1. Choose the boys, ignoring girls: $(2 5) = \frac{5 \cdot 4}{2} = 10.$ Why this step? The boy-choice is a self-contained combination; order among boys is irrelevant.

Step 2. Choose the girls, independently: $(3 6) = \frac{6 \cdot 5 \cdot 4}{6} = 20.$ Why this step? Same logic for girls, and this choice does not interact with which boys were picked.

Step 3. Combine by the multiplication principle: $(2 5) \cdot (3 6) = 10 \cdot 20 = 200.$ Why this step? For each of the 10 boy-groups, all 20 girl-groups are possible, so total pairs $= 10 \times 20$ . We multiply (independent stages), unlike Example 5 where disjoint cases were added.

Verify: Units sanity — we wanted committees, and $10$ (boy-ways) $\times$ $20$ (girl-ways) $= 200$ full committees. Every committee has exactly 2 boys and 3 girls as required, none double-counted. ✓

Example 7 — Cell I: a real-world word problem

Worked example Pizza toppings

A shop offers 8 distinct toppings. A "supreme" pizza uses exactly 3 of them, and toppings can't repeat. How many different supreme pizzas exist? Does adding pepperoni first vs last matter?

Forecast: guess whether it's under or over $60$ .

Step 1. Translate English → numbers: "8 distinct toppings" ⇒ $n = 8$ ; "exactly 3, no repeats" ⇒ $r = 3$ ; "a pizza doesn't remember the order you sprinkled toppings" ⇒ order irrelevant ⇒ combination. Why this step? The phrase "order doesn't matter" is the whole decision between $(3 8)$ and $^{8} P_{3}$ . A pizza with {ham, olive, chili} is identical to {chili, olive, ham}.

Step 2. Compute: $(3 8) = \frac{8 \cdot 7 \cdot 6}{3 !} = \frac{336}{6} = 56.$ Why this step? Top three falling factors over $3!$ — the standard un-shuffle.

Verify: Symmetry cross-check $(3 8) = (5 8)$ ; and Pascal's rule $(3 8) = (2 7) + (3 7) = 21 + 35 = 56.$ ✓ Two independent checks agree.

Example 8 — Cell J: the exam trap (order secretly matters)

Worked example Same numbers, different question — spot the switch

From 8 runners, (a) how many ways to pick a group of 3 to advance to the final, and (b) how many ways to award gold, silver, bronze medals to 3 of them?

Forecast: will (a) and (b) be equal? If not, which is bigger and by what factor?

Step 1 (a) — group, no order. Advancing is a plain committee: $(3 8) = \frac{8 \cdot 7 \cdot 6}{3 !} = 56.$ Why this step? Whether runner X advances "1st-named" or "3rd-named" makes no difference — same trio advances. Order irrelevant ⇒ combination.

Step 2 (b) — medals, order matters. Gold ≠ silver ≠ bronze, so the arrangement of the same three runners now gives different outcomes. This is a permutation: $^{8} P_{3} = 8 \cdot 7 \cdot 6 = 336.$ Why this step? Here swapping who gets gold and who gets bronze is a genuinely different result — so we must not divide by $3!$ . This is exactly the trap in the parent note's "Error A," run in reverse.

Step 3 — see the relationship. The two answers differ by precisely the un-shuffle factor: $^{8} P_{3} = (3 8) \cdot 3! = 56 \cdot 6 = 336.$ Why this step? Every unordered trio (56 of them) can be medal-ordered in $3! = 6$ ways, giving $336$ . This is the master relation $(r n) = \frac{^{n} P _{r}}{r !}$ , read forwards.

Verify: $56 \cdot 6 = 336$ ✓, and $336/6 = 56$ ✓. The single test — "does swapping two chosen people change the answer?" — cleanly separates (a) from (b).

Active recall

Why is

(0 6) = 1

and not

0

There is exactly one way to choose nothing (the empty set); the formula gives

\frac{6 !}{0 ! 6 !} = 1

using

0! = 1

What is

(8 6)

and why?

0

— you cannot choose more items than exist;

(6 - 8)!

is undefined, signalling impossibility.

Fast way to compute

(48 50)

Use symmetry

(48 50) = (2 50) = \frac{50 \cdot 49}{2} = 1225

"At least one chemist" from a pool — strategy?

Total minus complement:

(4 8) - (4 5) = 70 - 5 = 65

2 boys and 3 girls — add or multiply the sub-counts?

Multiply (independent stages):

(2 5) (3 6) = 10 \cdot 20 = 200

Group of 3 vs gold/silver/bronze from 8 — which is bigger and by how much?

Medals bigger by factor

3!

^{8} P_{3} = 336 = 6 \cdot (3 8) = 6 \cdot 56

Connections

Combinations — nCr, Pascal's triangle — the parent machinery these examples exercise.
Permutations — nPr — Example 8's trap lives on the order-matters side.
Factorials and 0! — Example 2's edge cases hinge on $0! = 1$ .
Binomial Theorem — the same $(r n)$ appear as coefficients of $(a + b)^{n}$ .
Probability — counting outcomes — these counts become numerators/denominators of probabilities.
Pascal's Triangle patterns — Example 4's addition law is the triangle's growth rule.