Permutations — nPr, arrangements with restrictions: Question bank

First — pin down the symbol

Everything on this page uses $n P_{r}$ . Before you trap-test yourself, make sure you can read it two ways.

The block (glue) and gap methods used throughout are also visual. Keep these two pictures in mind:

True or false — justify

Every answer below is a full reason. A bare "true/false" means you have not understood the trap.

A B

and

B A

count as two different permutations.

True. A permutation is an ordered arrangement, so swapping two items gives a genuinely new line-up — same letters, different order.

A B

and

B A

count as two different combinations.

False. A combination is a set (order ignored);

{A, B}

and

{B, A}

are the same selection, so they count once — see Combinations — nCr.

n P_{r}

always equals

n C_{r}

False.

n P_{r} = r! \cdot n C_{r}

(choose the

r

objects, then order them), so they are equal only when

r! = 1

, i.e.

r = 0

r = 1

. Otherwise

n P_{r}

is larger because it also counts the

r!

orderings.

n P_{n} = n!

relies on

0! = 1

True.

n P_{n} = \frac{n !}{( n - n )!} = \frac{n !}{0 !}

, and this only gives

n!

because

0!

is defined as

1

— see Factorials and 0!.

n P_{0} = 0

because you arrange nothing.

False.

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

: there is exactly one way to arrange nothing (the empty arrangement), just as there is one empty photo line-up.

For the "together" restriction, gluing two objects gives the final answer directly.

False. After treating the pair as one block you must still multiply by the block's internal

2!

orders, because the two glued objects can swap inside the block (see the glue figure above).

The gaps method and the "total − together" method give different answers for "must be apart".

False. They must agree — they count the same set of arrangements two ways. If they differ, you miscounted (usually forgot the internal

r!

in "together").

Arranging

k

objects in a row always creates exactly

k

gaps to slot new items into.

False. It creates

k + 1

gaps: one between each adjacent pair plus one at each end, e.g. _C_D_E_ gives

4

gaps for

k = 3

(see the gap figure above).

5 P_{5}

and

5 P_{4}

are equal.

True. In the falling-product form,

5 P_{4} = 5 \cdot 4 \cdot 3 \cdot 2

and

5 P_{5} = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1

; the extra last factor is

n - r + 1 = 5 - 5 + 1 = 1

, which multiplies by

1

and changes nothing. Both equal

120

You can use

n P_{r}

when repetition of objects is allowed.

False.

n P_{r}

assumes distinct, no-repeat picks — that is exactly why the falling product shrinks

n, n - 1, n - 2, \dots

. If repeats are allowed the pool stays at

n

every step, giving

n^{r}

instead — see Permutations with Repetition.

Why questions

Why do we multiply the choices at each position instead of adding them?

Because filling position 1 and then position 2 and then … is a chain of dependent decisions; "and-then" decisions multiply — this is the Fundamental Counting Principle. Picture a tree: each of the

n

first-slot branches sprouts

n - 1

second-slot branches, so the leaves multiply. Adding counts "either/or" alternatives, not sequential steps.

Why fill the most restricted position first?

If you spend freedom on the easy slots first, the tight slot may end up with zero valid choices left. In the even-number example, choosing hundreds and tens first can leave only odd digits for units. Locking the constraint first (units = even) guarantees it can always be satisfied.

Why does the "together" method glue objects into one block?

Look at the glue figure: a block that must stay adjacent behaves like a single unit for the between-object ordering; gluing turns "

k

objects with an adjacency rule" into "

k - 1

ordinary objects." We then correct for the two faces the block can show by multiplying by the internal

r!

Why does the "apart" case often use the complement (total − together)?

"Never together" is the exact opposite of "together." Every arrangement is either together or apart with nothing in between, so subtracting the together-count from the total leaves precisely the apart-count — usually easier than the direct gap-count, and the two must agree.

Why is

n P_{r}

written as

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

rather than the falling product

n (n - 1) \dots (n - r + 1)

They are the same number. Multiplying the falling product by

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

completes it into the full factorial

n!

on top while the untouched tail

(n - r)!

sits on the bottom and cancels the part you never used. The ratio is just a compact name for the shrinking product you actually compute.

Why does an extra constraint usually reduce the count?

A restriction removes valid choices at some step (a smaller factor in the product), and multiplying by a smaller factor yields a smaller product. It can never increase the arrangement count — at best (a constraint that forbids nothing) it leaves it unchanged.

Why can permutations act as the denominator in a probability?

When every arrangement is equally likely, the total number of arrangements is the size of the sample space, which sits under the count of favourable arrangements — see Probability — Equally Likely Outcomes.

Edge cases

What is

n P_{r}

when

r > n

It is

0

: the falling product

n (n - 1) \dots

eventually hits the factor

0

(you run out of objects), so the whole product is zero — you cannot arrange more distinct objects than you have.

What is

n P_{1}

Exactly

n

— the falling product has a single factor

n

; arranging one object into one slot is just choosing which of the

n

objects to place, with no order to worry about.

What is

0 P_{0}

1

, since

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

— the single empty arrangement, consistent with

n P_{0} = 1

If all

n

objects are identical, how many distinct permutations are there?

Just

1

: with no distinguishable objects, every "arrangement" looks the same. The distinctness assumption behind the falling product has collapsed — see Permutations with Repetition.

For a multiset — say the letters of

M I S S I S S I P P I

(11 letters, with repeats) — is

n P_{r}

the right tool?

No.

n P_{r}

assumes all objects are distinct. With repeated objects you first pretend they are distinct (

11!

) and then divide out the over-count for each repeated group:

\frac{11 !}{4 ! 4 ! 2 !}

(for the four

S

, four

I

, two

P

). This is the general rule the "identical pair" case is just the smallest instance of — see Permutations with Repetition.

For "at least one vowel at an end," is direct counting or complement better?

Complement is usually cleaner: count total arrangements, subtract those with no vowel at either end, since "at least one" spans many overlapping cases that are easy to double-count directly.

If two required-together objects are identical, do you still multiply by

2!

inside the block?

No. Swapping identical objects produces no new arrangement, so the internal factor is

\frac{2 !}{2 !} = 1

; multiplying by

2!

would over-count — the same multiset correction as

M I S S I S S I P P I

, just with a group of size

2

Flashcards

Two readings of

n P_{r}

The falling product

n (n - 1) \dots (n - r + 1)

(

r

shrinking factors) and the ratio

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

— the same number.

Why is

n P_{r} = r! \cdot n C_{r}

Build an ordered arrangement by choosing

r

objects (

n C_{r}

) then ordering them (

r!

); choose-and-then-order multiplies.

n P_{r}

larger or smaller than

n C_{r}

(for

r \geq 2

Larger, by a factor of

r!

, because permutations also count the orderings that combinations ignore.

What is

n P_{r}

when

r > n

0

— the falling product hits a zero factor; you cannot arrange more distinct objects than exist.

Gaps created by

k

arranged objects?

k + 1

(between each pair plus both ends).

Permutations of a multiset (repeated objects)?

\frac{n !}{n _{1} ! n _{2} ! \dots}

— full factorial divided by the factorial of each repeated group's size.

Do you multiply by internal

2!

when the glued pair is identical?

No — identical objects give no new order, so the factor is

1

Connections

Permutations — nPr, arrangements with restrictions — the parent topic these traps target.
Fundamental Counting Principle — why we multiply, not add.
Combinations — nCr — the "order ignored" contrast; source of the $n P_{r} = r! \cdot n C_{r}$ bridge.
Factorials and 0! — the $0! = 1$ edge cases.
Circular Permutations — the round-table trap.
Permutations with Repetition — multiset and identical-object traps.
Probability — Equally Likely Outcomes — where these counts become denominators.

Question bank — Permutations — nPr, arrangements with restrictions

First — pin down the symbol

True or false — justify

Spot the error

Why questions

Edge cases

Flashcards

Connections