Chapter 6 · Permutations and Combinations

6.2 Permutations

A permutation is an arrangement in a definite order. Filling $r$ places from $n$ different objects gives ${}^nP_r=\dfrac{n!}{(n-r)!}$ arrangements; when some objects are alike, divide out the swaps that change nothing.

Before a match, the captain of a cricket team writes down a batting order for the eleven players. How many different batting orders are possible? By the counting principle of Section 6.1, the first place can be filled in $11$ ways, the second in $10$, the third in $9$, and so on down to $1$: the answer is $11\times10\times9\times\cdots\times2\times1=39\,916\,800$. Almost forty million orders, from eleven names.

Products of consecutive whole numbers like this turn up in every arrangement problem, so they get a short name — the factorial. In this section we introduce factorial notation, count arrangements of $r$ objects chosen from $n$ different objects — the permutations ${}^nP_r$ — prove the formula ${}^nP_r=\dfrac{n!}{(n-r)!}$, and then deal with the common case where some of the objects are identical, like the three A's in BANANA. Along the way you will learn the two standard tricks for "together" and "never together".

Factorial Notation

📘 Definition 6.2 — Factorial

For a natural number $n$, the product of the first $n$ natural numbers is called $n$ factorial and written $n!$: $$n!=1\times2\times3\times\cdots\times(n-1)\times n.$$ We also define $$0!=1.$$ So $1!=1$, $2!=2$, $3!=6$, $4!=24$, $5!=120$, $6!=720$, $7!=5040$.

The most useful property is that each factorial contains the one before it: $$n!=n\times(n-1)!\qquad(n\ge1),$$ and, peeling off more factors, $n!=n(n-1)(n-2)!$ for $n\ge2$, and so on. This is what makes quotients of factorials easy: $\dfrac{8!}{6!}=\dfrac{8\times7\times6!}{6!}=56$ — never multiply out $8!$ and $6!$ separately.

Why is $0!=1$ and not $0$? Put $n=1$ in $n!=n\times(n-1)!$: it says $1!=1\times0!$, so $0!$ must be $1$ if the rule is to keep working. In counting terms, there is exactly one way to arrange nothing — do nothing. You will see in Theorem 6.4 that this choice makes the formulas work without exceptions.

✏️ Example 1 — Working with factorials

(a) Evaluate $5!$ and $7!-5!$.
(b) Evaluate $\dfrac{8!}{6!\,2!}$.
(c) Find the natural number $n$ with $(n+1)!=12\,(n-1)!$.
(d) Find $x$ if $\dfrac{1}{6!}+\dfrac{1}{7!}=\dfrac{x}{8!}$.

💡 Show solution▶

(a) $5!=120$. $7!-5!=5040-120=4920$. (Note: $7!-5!\ne2!$ — factorials do not subtract like that.)

(b) $\dfrac{8\times7\times6!}{6!\times2}=\dfrac{56}{2}=28$.

(c) $(n+1)!=(n+1)\,n\,(n-1)!$, so $(n+1)n=12$, i.e. $n^2+n-12=0$, $(n+4)(n-3)=0$. Since $n$ is natural, $n=3$. Check: $4!=24=12\times2!$ ✓.

(d) Multiply both sides by $8!$: $\dfrac{8!}{6!}+\dfrac{8!}{7!}=x$, so $x=56+8=64$.

Permutations of Different Objects

📘 Definition 6.3 — Permutation

A permutation is an arrangement, in a definite order, of some or all of a given collection of objects. For $0\le r\le n$, the number of permutations of $n$ different objects taken $r$ at a time, with no object repeated, is written $${}^nP_r.$$ For example, the permutations of the letters A, B, C taken $2$ at a time are $\text{AB},\text{BA},\text{AC},\text{CA},\text{BC},\text{CB}$, so ${}^3P_2=6$.

📐 Theorem 6.4 — Formula for ${}^nP_r$

For integers $0\le r\le n$, $$\begin{aligned}{}^nP_r&=n(n-1)(n-2)\cdots(n-r+1)\\&=\dfrac{n!}{(n-r)!}.\end{aligned}$$ In particular ${}^nP_n=n!$ and ${}^nP_0=1$.

Proof. Arranging $r$ of the objects means filling $r$ places in order. The first place can be filled by any of the $n$ objects. Whichever object is used, $n-1$ objects are left for the second place, $n-2$ for the third, …, and for the $r$-th place $n-(r-1)=n-r+1$ objects are left. By the fundamental principle of counting (Theorem 6.2), $${}^nP_r=n(n-1)\cdots(n-r+1).$$ Multiply and divide by $(n-r)!=(n-r)(n-r-1)\cdots2\cdot1$: the numerator becomes $n(n-1)\cdots(n-r+1)(n-r)\cdots1=n!$, so ${}^nP_r=\dfrac{n!}{(n-r)!}$. For $r=n$ this gives $\dfrac{n!}{0!}=n!$, and for $r=0$ it gives $\dfrac{n!}{n!}=1$ (one way to arrange nothing) — both correct only because $0!=1$. $\blacksquare$

📐 Theorem 6.5 — Permutations with repetition

The number of arrangements of $r$ objects taken from $n$ different objects, when each object may be used any number of times, is $$n^r.$$ Reason. Each of the $r$ places can be filled in $n$ ways whatever was put in the other places (the Corollary of Theorem 6.2). $\blacksquare$

✏️ Example 2 — Words from MONDAY

How many words, with or without meaning, can be made from the letters of $\text{MONDAY}$, using each letter at most once, if (a) $4$ letters are used (b) all the letters are used (c) all the letters are used and the first letter is a vowel?

💡 Show solution▶

(a) $6$ different letters, $4$ places: ${}^6P_4=6\times5\times4\times3=360$.

(b) ${}^6P_6=6!=720$.

(c) The vowels are O and A. First place: $2$ ways; the other $5$ letters fill the remaining $5$ places in $5!=120$ ways. Total $2\times120=240$.

Check (c) By symmetry, each of the $6$ letters starts $\tfrac{720}{6}=120$ of the words in (b); two of them are vowels: $240$ ✓.

✏️ Example 3 — Equations in ${}^nP_r$

(a) Find $n$ if ${}^nP_4=20\times{}^nP_2$.
(b) Find $r$ if ${}^5P_r=2\times{}^6P_{r-1}$.

💡 Show solution▶

(a) $n(n-1)(n-2)(n-3)=20\,n(n-1)$. Here $n\ge4$, so $n(n-1)\ne0$ and we may divide: $(n-2)(n-3)=20$, $n^2-5n-14=0$, $(n-7)(n+2)=0$, so $n=7$.

(a) Check ${}^7P_4=840$ and $20\times{}^7P_2=20\times42=840$ ✓.

(b) $\dfrac{5!}{(5-r)!}=2\cdot\dfrac{6!}{(7-r)!}$. Multiply by $(7-r)!$ and use $(7-r)!=(7-r)(6-r)(5-r)!$: $5!\,(7-r)(6-r)=2\cdot6!$, so $(7-r)(6-r)=12$, $r^2-13r+30=0$, $(r-3)(r-10)=0$.

(b) Answer $r=10$ is impossible (we need $r\le5$), so $r=3$. Check: ${}^5P_3=60=2\times{}^6P_2=2\times30$ ✓.

🧭 Problem-solving strategy — "Together" and "never together"

Together (block method). Tie the objects that must be together into one block. Arrange the block with the other objects as single units, then multiply by the number of ways of arranging the objects inside the block.
Never together, two objects. Subtract: $\#(\text{never together})=\#(\text{all})-\#(\text{together})$.
No two of a group together (gap method). First arrange the other objects. They create gaps — one before each object and one at the end. Choose and fill gaps for the group, one per gap.

✏️ Example 4 — Together and apart

$5$ boys and $3$ girls are to stand in a row for a photograph. In how many ways can they stand if (a) there is no restriction (b) the three girls stand together (c) no two girls stand next to each other?

💡 Show solution▶

(a) $8$ different people: $8!=40\,320$ ways.

(b) Block Tie the $3$ girls into one block. The block and $5$ boys are $6$ units: $6!=720$ arrangements. Inside the block the girls can stand in $3!=6$ orders. Total $720\times6=4320$.

(c) Gaps Arrange the $5$ boys: $5!=120$ ways. This makes $6$ gaps: $\_\,\text{B}\,\_\,\text{B}\,\_\,\text{B}\,\_\,\text{B}\,\_\,\text{B}\,\_$. Put the $3$ girls into $3$ different gaps, in order: ${}^6P_3=120$ ways. Total $120\times120=14\,400$.

Check Every arrangement either has no two girls adjacent or has some adjacent pair. The adjacent cases number $40\,320-14\,400=25\,920$, which is more than the $4320$ in (b) because "some two girls together" includes arrangements with only two of the three side by side. (A computer listing of all $8!$ arrangements confirms $14\,400$.)

Permutations When Some Objects Are Alike

How many different words can be made from the letters of $\text{BOOK}$? If the two O's were different — say $\text{O}_1$ and $\text{O}_2$ — there would be $4!=24$ arrangements. But $\text{BO}_1\text{O}_2\text{K}$ and $\text{BO}_2\text{O}_1\text{K}$ both read $\text{BOOK}$: swapping the two O's changes nothing you can see. Every visible word is counted $2!=2$ times among the $24$, so there are $\tfrac{24}{2}=12$ words.

📐 Theorem 6.6 — Permutations of objects not all different

Suppose $n$ objects consist of $p_1$ alike objects of one kind, $p_2$ alike objects of a second kind, …, $p_k$ alike objects of a $k$-th kind, and the rest (if any) all different from each other and from these. Then the number of permutations of all $n$ objects is $$\dfrac{n!}{p_1!\,p_2!\cdots p_k!}.$$

Proof. Let the number of permutations be $x$. Take any one of them and make the alike objects different by attaching labels: $\text{O}_1,\text{O}_2$, and so on. Keeping every other object where it is, the $p_1$ labelled objects of the first kind can be rearranged among the positions they occupy in $p_1!$ ways, those of the second kind in $p_2!$ ways, and so on. By the fundamental principle, each of the $x$ permutations produces $p_1!\,p_2!\cdots p_k!$ permutations of $n$ different objects. Every permutation of the $n$ different (labelled) objects arises exactly once in this way — remove its labels to see which of the $x$ it comes from. Therefore $$x\cdot p_1!\,p_2!\cdots p_k!=n!,$$ and dividing gives the formula. $\blacksquare$

✏️ Example 5 — The letters of COMMITTEE

Find the number of arrangements of all the letters of $\text{COMMITTEE}$. How many of them (a) begin and end with E (b) have all the vowels together?

💡 Show solution▶

Count letters $9$ letters: C, O, M, M, I, T, T, E, E — that is M twice, T twice, E twice, and C, O, I once. Total $\dfrac{9!}{2!\,2!\,2!}=\dfrac{362\,880}{8}=45\,360$.

(a) Fix E at both ends: $\text{E}\,\_\,\_\,\_\,\_\,\_\,\_\,\_\,\text{E}$. The middle holds C, O, M, M, I, T, T: $\dfrac{7!}{2!\,2!}=1260$.

(b) Block The vowels O, I, E, E form one block. The units are the block and C, M, M, T, T: $6$ units with M and T each twice: $\dfrac{6!}{2!\,2!}=180$. Inside the block: $\dfrac{4!}{2!}=12$ orders.

(b) Answer $180\times12=2160$. Notice that the "alike" division is needed both outside the block (M, T) and inside it (E).

✏️ Example 6 — Digits that repeat

How many 7-digit numbers can be made using all of the digits $1,2,2,3,3,3,4$? How many of them are odd?

💡 Show solution▶

All $7$ digits, $2$ appearing twice and $3$ three times: $\dfrac{7!}{2!\,3!}=\dfrac{5040}{12}=420$. (No $0$, so there is no first-digit problem.)

Odd: cases The last digit must be $1$ or $3$, and the alike-counts left over differ, so split. Last digit $1$: arrange $2,2,3,3,3,4$: $\dfrac{6!}{2!\,3!}=60$. Last digit $3$: arrange $1,2,2,3,3,4$: $\dfrac{6!}{2!\,2!}=180$.

Answer $60+180=240$ odd numbers.

Check The even ones end in $2$ or $4$: $\dfrac{6!}{3!}+\dfrac{6!}{2!\,3!}=120+60=180$, and $240+180=420$ ✓.

Pick a word, or type your own (up to $6$ letters). The lab lists every different arrangement of its letters, one by one, and counts them. Below the list, the highlighted word is shown with its repeated letters labelled ($\text{A}_1,\text{A}_2,\dots$): all those labelled versions look the same once the labels are removed. The number of copies is always $p_1!\,p_2!\cdots$, so the count of different words is $\dfrac{n!}{p_1!\,p_2!\cdots}$.

word

listed so far—

$n!$ (labelled)—

copies per word—

different words—

⚠️ Common mistakes with permutations

1. Confusing ${}^nP_r$ with $n^r$. ${}^nP_r$ forbids repetition; $n^r$ allows it. "Using each letter at most once" means ${}^nP_r$.

2. $0!=0$. No: $0!=1$. With $0!=0$ the formula ${}^nP_n=\dfrac{n!}{0!}$ would divide by zero.

3. Forgetting to divide for alike objects. The letters of $\text{LEVEL}$ give $\dfrac{5!}{2!\,2!}=30$ words, not $5!=120$. Count every repeated letter, and divide by the factorial of each repetition count.

4. Forgetting the arrangements inside a block. In "the girls stand together", after arranging the units you must still multiply by the number of orders inside the block.

5. Cancelling factorials wrongly. $\dfrac{8!}{4!}\ne2!$ and $\dfrac{8!}{4!}\ne4!$; it is $8\times7\times6\times5=1680$. Similarly $(n+1)!\ne n!+1$.

🚀 Beyond the board syllabus (JEE Main · JEE Advanced)

In JEE Main and JEE Advanced 2026 (both list permutations and their applications); not in the CBSE year-end exam. Circular arrangements are a favourite of entrance-exam papers.

Circular permutations. When $n$ different people sit round a circular table, only their positions relative to each other matter: rotating everyone one seat to the left gives the same seating. Each circular seating corresponds to $n$ row arrangements (cut the circle at any of the $n$ places), so the number of circular arrangements is $$\dfrac{n!}{n}=(n-1)!.$$ Equivalently, fix one person's seat and arrange the other $n-1$ in a row. If, in addition, clockwise and anticlockwise orders count as the same — beads on a necklace, flowers on a garland — the number is $\dfrac{(n-1)!}{2}$ for $n\ge3$.

Worked example. In how many ways can $6$ people sit at a round table? In how many of these do two particular people, P and Q, sit next to each other?
Solution. $(6-1)!=120$ ways. For P and Q together, treat them as one unit: $5$ units round the table give $(5-1)!=24$ arrangements, and P, Q can sit in $2$ orders within the unit: $48$. Check: fix P's seat; Q has $5$ seats to choose from, $2$ of them next to P, so P and Q are neighbours in $\tfrac25$ of all seatings, and $\tfrac25\times120=48$ ✓.

Exercises

Factorials.

1 Evaluate $6!$.
2 Evaluate $\dfrac{8!}{5!\,3!}$.
3 Is $3!+4!=7!$? Evaluate both sides.
4 Evaluate $\dfrac{n!}{r!\,(n-r)!}$ for $n=7$, $r=3$.
5 Find $x$ if $\dfrac{1}{8!}+\dfrac{1}{9!}=\dfrac{x}{10!}$.
6 Find the natural number $n$ with $(n+2)!=2550\times n!$.
7 Find $n$ if $\dfrac{n!}{(n-2)!}=56$.
8 Write $10\times9\times8\times7$ as a quotient of two factorials.

💡 Show solutions▶

1 $720$.

2 $\dfrac{8\times7\times6}{3!}=\dfrac{336}{6}=56$.

3 No: $3!+4!=6+24=30$, while $7!=5040$.

4 $\dfrac{5040}{6\times24}=35$.

5 Multiply by $10!$: $x=\dfrac{10!}{8!}+\dfrac{10!}{9!}=90+10=100$.

6 $(n+2)(n+1)=2550=51\times50$, so $n+1=50$, $n=49$ (the other root, $n=-52$, is not natural).

7 $n(n-1)=56=8\times7$, so $n=8$.

8 $\dfrac{10!}{6!}$ (which is ${}^{10}P_4$).

Permutations of different objects.

1 Evaluate ${}^7P_3$.
2 Evaluate ${}^{10}P_2$.
3 Evaluate ${}^6P_6$.
4 Find $n$ if ${}^nP_2=72$.
5 Find $n$ if ${}^nP_5=42\times{}^nP_3$ ($n\ge5$).
6 Find $r$ if ${}^6P_r=120$.
7 How many 3-digit numbers with three different digits can be made from the digits $1$–$9$?
8 In how many ways can $4$ people sit on a row of $6$ chairs?

💡 Show solutions▶

1 $7\times6\times5=210$.

2 $10\times9=90$.

3 $6!=720$.

4 $n(n-1)=72=9\times8$, so $n=9$.

5 Divide by ${}^nP_3=n(n-1)(n-2)\ne0$: $(n-3)(n-4)=42=7\times6$, so $n=10$. Check: ${}^{10}P_5=30\,240=42\times720$ ✓.

6 $6\times5\times4=120$, so $r=3$.

7 ${}^9P_3=504$.

8 The people choose chairs in order: ${}^6P_4=6\times5\times4\times3=360$.

III

Restrictions and alike objects.

1 How many arrangements of the letters of $\text{DELHI}$ are there, and how many of them begin with a vowel?
2 How many arrangements of the letters of $\text{PATNA}$ are there?
3 How many arrangements of the letters of $\text{ASSESS}$ are there?
4 How many arrangements of the letters of $\text{INSTITUTE}$ are there? In how many are the vowels together?
5 $7$ different books are put on a shelf. In how many ways can this be done if two particular books are (a) together (b) never together?
6 In how many ways can $4$ boys and $4$ girls stand in a row so that boys and girls alternate?
7 In how many ways can $3$ identical red balls, $2$ identical blue balls and $1$ green ball be placed in a row?
8 How many 6-digit numbers can be made using all of the digits $0,1,1,2,2,2$?
9 A signal is made by hanging all of $6$ flags in a vertical line: $2$ identical red, $2$ identical blue and $2$ identical green flags. How many signals are possible?

Check: count the arrangements of any word by listing them

letters

formula $n!/(p_1!p_2!\cdots)$—

by listing every arrangement—

💡 Show solutions▶

1 $5!=120$. Vowels E, I: $2$ choices for the first letter, then $4!$: $2\times24=48$.

2 A twice: $\dfrac{5!}{2!}=60$.

3 S four times: $\dfrac{6!}{4!}=30$.

4 I twice, T three times: $\dfrac{9!}{2!\,3!}=30\,240$. Vowels I, I, U, E as a block with N, S, T, T, T: $\dfrac{6!}{3!}=120$ arrangements of units, $\dfrac{4!}{2!}=12$ inside the block: $1440$.

5 (a) Block: $6!\times2!=1440$. (b) $7!-1440=5040-1440=3600$.

6 Pattern BGBGBGBG or GBGBGBGB ($2$ ways); boys in their places $4!$, girls $4!$: $2\times24\times24=1152$.

7 $\dfrac{6!}{3!\,2!}=60$.

8 All arrangements: $\dfrac{6!}{2!\,3!}=60$. Those starting with $0$: arrange $1,1,2,2,2$ after it: $\dfrac{5!}{2!\,3!}=10$. Numbers: $60-10=50$.

9 $\dfrac{6!}{2!\,2!\,2!}=90$.

Applications.

1 A cricket captain fixes the batting order of his $11$ players. The two opening places must go to two of the $3$ specialist openers in the team; the other $9$ players (including the third opener) fill places $3$ to $11$ in any order. How many batting orders are possible?
2 A railway line has $20$ stations. A ticket shows the starting station and the destination. How many different kinds of single-journey ticket must be printed?
3 A suitcase lock has $4$ rings, each showing the $6$ letters A–F. What is the largest number of unsuccessful attempts someone who has forgotten the code could make?
4 In a race with $8$ runners, in how many ways can the gold, silver and bronze medals be awarded (no ties)?
5 A DNA strand is a sequence of the bases A, C, G, T. How many different strands of length $10$ are possible?
6 In how many arrangements of the letters of $\text{MATHS}$ are M and A not next to each other?
7 Five board papers, one of them Mathematics, are to be held on five different days. In how many orders can they be scheduled if Mathematics is not on the first day?

💡 Show solutions▶

1 Openers in order: ${}^3P_2=6$; the other $9$ places: $9!=362\,880$. Total $6\times362\,880=2\,177\,280$.

2 Ordered pairs of different stations: ${}^{20}P_2=380$ tickets.

3 Codes: $6^4=1296$ (repetition allowed); at most $1295$ attempts fail.

4 ${}^8P_3=336$.

5 $4^{10}=1\,048\,576$.

6 All: $5!=120$. M, A together: $4!\times2=48$. Not together: $72$.

7 All orders $5!=120$; Mathematics first: $4!=24$; answer $96$. (Or: $4$ choices of day for Mathematics, then $4!$: $4\times24=96$ ✓.)

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE MainJEE Advanced

How this is examined: Board papers ask you to evaluate factorials, solve an equation in ${}^nP_r$ (2 marks) and count arrangements of a word with repeated letters, often with a "begins with" or "together" condition (3–5 marks). JEE adds dictionary ranks, circular arrangements and identities between the ${}^nP_r$.

E1. Board · 1 mark The value of $0!+1!+2!$ is (A) $3$ (B) $4$ (C) $2$ (D) $0$

💡 Show solution▶

Idea $0!=1$, $1!=1$, $2!=2$; the sum is $4$.

Trap (A) comes from taking $0!=0$.

Answer (B).

E2. Board · 2 marks Find $n$ if ${}^nP_4:{}^nP_2=12:1$.

💡 Show solution▶

Set up $\dfrac{n(n-1)(n-2)(n-3)}{n(n-1)}=12$, so $(n-2)(n-3)=12$. (1 mark)

Answer $n^2-5n-6=0$, $(n-6)(n+1)=0$, and $n\ge4$, so $n=6$. Check: ${}^6P_4=360$, ${}^6P_2=30$, ratio $12$ ✓. (1 mark)

E3. Board · 3 marks Find the number of arrangements of the letters of the word $\text{PARALLEL}$. In how many of them (i) does the word begin with P (ii) are all three L's together?

💡 Show solution▶

Total $8$ letters: P, R, E once; A twice; L three times. $\dfrac{8!}{2!\,3!}=\dfrac{40\,320}{12}=3360$. (1 mark)

(i) P fixed first; arrange A, A, R, L, L, L, E: $\dfrac{7!}{2!\,3!}=420$. (1 mark)

Answer (ii) Block LLL (only one internal order, since the L's are alike) with P, A, A, R, E: $6$ units, A twice: $\dfrac{6!}{2!}=360$. (1 mark)

E4. Board · Case study Six students, including Asha and Bilal, are to sit in a row of six chairs for a group photograph. (i) In how many ways can they sit? (1 mark) (ii) In how many ways if Asha and Bilal sit together? (1 mark) (iii) In how many ways if Asha and Bilal do not sit together, and in how many of those does Asha sit at one end of the row? (2 marks)

💡 Show solution▶

(i) $6!=720$. (1 mark)

(ii) Block: $5!\times2!=240$. (1 mark)

(iii) first part $720-240=480$. (1 mark)

Answer (iii) Asha at an end: $2$ choices of end. Bilal must avoid the chair next to her: $4$ choices. The other $4$ students: $4!=24$. Total $2\times4\times24=192$ (out of the $480$). (1 mark)

E5. JEE Main All the arrangements of the letters of $\text{TRAIN}$ are listed in dictionary (alphabetical) order. The position of the word $\text{TRAIN}$ in the list is (A) $113$ (B) $114$ (C) $115$ (D) $116$

💡 Show solution▶

Idea Alphabetical order of the letters: A, I, N, R, T. Count the words that come before TRAIN, letter by letter.

Count Words starting with A, I, N or R: $4\times4!=96$. Starting with T, second letter A, I or N (all before R): $3\times3!=18$. Starting TR: the next letters A, I, N are already in alphabetical order, so TRAIN is the first word beginning with TR.

Trap $96+18=114$ words come before TRAIN; forgetting to add $1$ gives (B).

Answer (C) $114+1=115$.

E6. JEE Advanced One or more options correct. For integers $1\le r\le n$: (A) ${}^nP_r=n\cdot{}^{n-1}P_{r-1}$ (B) ${}^nP_r=(n-r+1)\cdot{}^nP_{r-1}$ (C) ${}^nP_n={}^nP_{n-1}$ (D) ${}^nP_r={}^nP_{n-r}$

💡 Show solution▶

(A) True: $n\cdot\dfrac{(n-1)!}{(n-r)!}=\dfrac{n!}{(n-r)!}$. In words: choose the object for the first place ($n$ ways), then arrange $r-1$ of the remaining $n-1$.

(B) True: $(n-r+1)\cdot\dfrac{n!}{(n-r+1)!}=\dfrac{n!}{(n-r)!}$. Fill $r-1$ places, then the $r$-th place has $n-r+1$ choices.

(C), (D) (C) True: $\dfrac{n!}{0!}=\dfrac{n!}{1!}$ — once $n-1$ places are filled, the last object has only one place to go. (D) False: ${}^4P_1=4$ but ${}^4P_3=24$.

Answer (A), (B), (C).

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. Whose name in your class can be rearranged in the most ways — and how much do repeated letters cut the number down?

Why it matters. The formula $\dfrac{n!}{p_1!\,p_2!\cdots}$ is used far beyond word puzzles: chemists count the arrangements of molecules in a crystal with it, biologists count DNA sequences with a given make-up, and it is the starting point of the idea of entropy in physics — the more arrangements a state has, the more likely it is.

What to do.

Collect the first names of $15$–$20$ classmates (with permission). For each name write $n$, the repeated letters and their counts.
Compute the number of arrangements $\dfrac{n!}{p_1!\,p_2!\cdots}$ for every name in a spreadsheet (use FACT()).
For one short name with a repeated letter, list all its arrangements by hand to check the formula.
Rank the names. Find two names of the same length whose counts differ a lot, and explain why using the formula.

Tools: A spreadsheet app; paper for the hand listing

Hand in: Your table, the ranking, the hand-made list, and a short paragraph on how repetition reduces the count.

Going further. A shuffled pack of $52$ different cards can be in $52!$ orders. Estimate $52!$ with a calculator and compare it with the age of the universe in seconds (about $4.4\times10^{17}$). Is any shuffle you make likely to have happened before?

Instructor note: e.g. PRIYA $5!=120$; ADITYA $\tfrac{6!}{2!}=360$; ANANYA (A three times, N twice) $\tfrac{6!}{3!\,2!}=60$ — same length as ADITYA but six times fewer. $52!\approx8.07\times10^{67}$, about $1.8\times10^{50}$ times the age of the universe in seconds, so a well-shuffled deck is almost certainly in an order never seen before.

🧪 Self-check quiz

0/5 correct

Q1. $0!$ is equal to

Q2. ${}^5P_2$ is

Q3. The number of different arrangements of the letters of $\text{LEVEL}$ is

Q4. In how many ways can $4$ people A, B, C, D stand in a row if A and B must stand next to each other?

Q5. How many 3-letter codes can be made from the letters A, B, C, D, E if letters may be repeated?

← 6.1 The Fundamental Principle of Counting 6.3 Combinations →