Combinations — nCr, Pascal's triangle: Question bank

Figure — Combinations — nCr, Pascal's triangle

Look at the triangle above: every valid $(r n)$ sits inside the two orange edges (which are all $1$ 's, the $r = 0$ and $r = n$ cases). Anything you'd write outside those edges is treated as $0$ — that's the visual meaning of " $r > n$ gives $0$ ".

True or false — justify

Each answer must contain a reason, not a verdict. Speak the "because" aloud. Assume $0 \leq r \leq n$ (both whole numbers) unless a statement says otherwise.

(r n)

and

^{n} P_{r}

count the same thing whenever

r = 1

True — choosing one item and arranging one item are identical, since a single item has only

1! = 1

ordering, so

(1 n) = \frac{^{n} P _{1}}{1 !} = n =^{n} P_{1}

(r n)

is always an integer even though the formula is a fraction.

True — it counts real committees, so it cannot be fractional; the

r! (n - r)!

in the denominator always divides

n!

exactly because

n!

contains the factors of every group's internal orderings.

(r n) = (n - r n)

only holds when

r < n - r

False — for every

r

in the valid range

0 \leq r \leq n

it holds; picking the

r

to keep always simultaneously names the

n - r

to discard, regardless of which is larger.

Increasing

n

by 1 always increases

(r n)

for a fixed

r

with

1 \leq r \leq n

True — provided

r \geq 1

, you gain new committees that include the new person while keeping all old ones, so the count strictly grows. (For

r = 0

it is false:

(0 n) = 1

for every

n

, so nothing changes.)

Increasing

r

by 1 always increases

(r n)

False — over the valid range

0 \leq r \leq n

the row rises then falls;

(r n)

peaks near

r = n /2

and shrinks back to

1

r = n

, so for

r

past the middle, raising

r

decreases the count.

The sum of every entry in row

n

of Pascal's triangle is

2^{n}

True — each of the

n

items is independently either in or out of a subset, giving

2^{n}

subsets total, which is exactly

\sum_{r = 0}^{n} (r n)

. See Pascal's Triangle patterns.

(r n)

counts ordered pairs of teams when you split

n

people into a group of

r

and a group of

n - r

False — it counts one unordered choice; the split into two groups is a single decision, which is precisely why

(r n) = (n - r n)

(the two names describe the same act).

For a valid combination, the

n

items must be distinct.

True — the derivation divides by

r!

assuming every internal arrangement is genuinely different; if items repeat, those "arrangements" aren't distinct and the count over-corrects.

Spot the error

Each line states a wrong claim or step. The answer names the flaw and repairs it.

"A committee of 3 from 7 gives

^{7} P_{3} = 210

choices."

Wrong tool — permutations count ordered selections, but a committee has no positions. Each single committee of 3 people can be listed in

3! = 6

different orders, so

^{7} P_{3}

counts every committee exactly

3!

times; divide by

3!

to erase that ordering, giving

(3 7) = 35

(0 n) = 0

because choosing nothing gives no committee."

The empty committee is a valid, single choice; there is exactly one way to pick nothing, so

(0 n) = \frac{n !}{0 ! n !} = 1

, using

0! = 1

from Factorials and 0!.

"In Pascal's triangle, each entry is the sum of its two neighbours in the same row."

The addition law adds the two entries diagonally above it, i.e.

(r - 1 n - 1) + (r n - 1)

from the previous row — not same-row neighbours.

(2 5) = \frac{5 !}{2 !} = 60

The denominator is missing

(n - r)! = 3!

; the correct formula is

\frac{5 !}{2 ! 3 !} = 10

— forgetting the

(n - r)!

factor is the most common slip.

(9 7)

is undefined so we write it as

(- 2 7)

When

r > n

you cannot choose more items than exist, so

(9 7) = 0

directly; there's no need to invent negative

r

— it's simply an impossible selection.

"From a club of 6 boys and 4 girls (10 people total), choose a 4-person team with at least one girl: just add up

(1 4) + (2 4) + (3 4) + (4 4)

and you're done."

That ignores which of the 6 boys fill the remaining seats, so it undercounts; the clean route is total minus complement — all teams minus all-boy teams —

(4 10) - (4 6) = 210 - 15 = 195

(see Probability — counting outcomes).

(98 100)

needs a hundred-digit factorial, so it's uncomputable by hand."

Use symmetry:

(98 100) = (2 100) = \frac{100 \cdot 99}{2} = 4950

— always convert to the smaller

r

before multiplying.

"Since order doesn't matter,

(2 4) = \frac{4 \cdot 3}{2} = 6

pairs, and each pair can be arranged 2 ways, so really there are 12."

The "arrange each pair 2 ways" step re-introduces the very ordering we removed; the whole point of dividing by

2!

was to treat those two arrangements as one — the answer stays

6

Why questions

The reasoning is the answer — a bare fact scores nothing.

Why do we divide

^{n} P_{r}

r!

and not by some other number?

Because every unordered group of size

r

appears once for each of its own internal orderings, and

r

items rearrange in exactly

r!

ways; dividing by

r!

collapses those duplicates into a single count.

Why is

(r n)

symmetric, i.e. why does choosing

r

equal choosing

n - r

Every time you select the

r

you keep, you have simultaneously and uniquely selected the

n - r

you reject — one action, two descriptions, so the two counts must be equal.

Why does Pascal's rule add the two entries above instead of multiplying or subtracting them?

Because fixing one special person splits all committees into two disjoint, exhaustive cases (that person in, or out); disjoint cases are combined by addition, never multiplication.

Why can we write only the top

r

falling factors, e.g.

7 \cdot 6 \cdot 5

, instead of full

7!

Those

r

factors are

^{n} P_{r} = \frac{n !}{( n - r )!}

, since the

(n - r)!

tail of

n!

cancels; writing the full factorial just wastes effort computing digits that immediately vanish.

Why do the binomial coefficients in

(a + b)^{n}

turn out to be exactly

(r n)

Expanding the product means choosing, from each of the

n

factors, either

a

b

; the number of ways to pick

b

from exactly

r

of the factors is

(r n)

, which becomes the coefficient of

a^{n - r} b^{r}

— see Binomial Theorem.

Why must the

n

items be distinct for the combination formula to hold?

The division by

r!

assumes all

r!

internal arrangements of a chosen group are genuinely different; if two items are identical, some "arrangements" coincide and dividing by the full

r!

removes too much.

Why is

(r n)

largest in the middle of a row rather than at an end?

There are far more ways to pick a balanced subset than a nearly-empty or nearly-full one — extremes are pinned to

1

(all-in or all-out), while middle sizes have vast freedom, so the count bulges centrally.

Edge cases

Boundary inputs where the formula must still behave. Test that you know the value and why.

What is

(n n)

, and why?

It is

1

— there is exactly one way to choose all

n

items (take everyone), and the formula gives

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

using

0! = 1

What is

(0 0)

1

— from an empty set there is exactly one subset, the empty set itself, and

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

What is

(r n)

when

r > n

0

— you cannot select more items than the set contains, so no such subset exists; the value is genuinely zero, not undefined.

What is

(0 5) + (5 5)

, and what does it represent?

1 + 1 = 2

— the two most extreme choices, "pick none" and "pick all", each achievable in exactly one way; they're the mirror ends of row 5.

(1 n) = n

obvious from the counting meaning?

Yes — choosing exactly one item from

n

distinct items gives

n

possibilities, one per item, with no ordering to worry about, and the formula confirms

\frac{n !}{1 ! ( n - 1 )!} = n

Does Pascal's rule still work at the edges of a row, like getting the leading

1

Yes if we treat off-triangle entries as

0

(0 n) = (- 1 n - 1) + (0 n - 1) = 0 + 1 = 1

, so the border of

1

's emerges naturally from the addition law.

What does

(r n)

become if

r

is negative?

We commit to the counting convention

(r n) = 0

for every

r < 0

— a subset cannot contain a negative number of elements, so no such selection exists and its count is zero; this choice is what keeps Pascal's rule and identities like the row-sum valid at the boundaries.

Recall One-line summary of every trap

The traps all reduce to three questions: Does order matter? (permutation vs combination), Did I keep the $(n - r)!$ and treat $0! = 1$ ? (formula boundaries), and Are my cases disjoint? (when to add vs multiply). Keep those three straight and no trap can catch you.

Connections

Combinations — $\binom{n}{r}$ and Pascal's Triangle — the parent note these traps stress-test.
Permutations — nPr — the "does order matter?" trap lives on this boundary.
Factorials and 0! — the $0! = 1$ convention resolves the empty-choice edge cases.
Binomial Theorem — why the coefficients are combinations.
Probability — counting outcomes — the "at least one" complement trick.
Pascal's Triangle patterns — row-sum $2^{n}$ and the addition-law direction traps.