A permutation is an arrangement of objects in a definite order. The instant the question cares about which object goes where — first place versus second, captain versus vice-captain, the order of letters in a word — you are counting permutations.
The number of permutations of $n$ distinct objects taken $r$ at a time is written ${}^{n}P_{r}$ (read "$n$ permute $r$"). By the multiplication principle, the first slot has $n$ choices, the second $n-1$, and so on for $r$ slots:
$${}^{n}P_{r} = n(n-1)(n-2)\cdots(n-r+1) = \dfrac{n!}{(n-r)!}, \quad 0 \le r \le n$$
Two special cases fall straight out of the formula. Arranging all $n$ objects gives ${}^{n}P_{n} = \dfrac{n!}{0!} = n!$, and arranging none gives ${}^{n}P_{0} = 1$. (Here lies the practical reason $0!$ must equal $1$: without it, the tidy formula ${}^{n}P_{n} = n!$ would break.)
Permutations with repetition allowed. If each of the $r$ positions may use any of the $n$ objects again, every slot independently has $n$ choices, so the count is simply $n^r$. This is the rule for things like PIN codes or number plates where digits may repeat.
$$\text{with repetition: } n^{r}$$
Permutations of objects not all distinct. When the collection contains repeated objects, swapping two identical objects produces the same arrangement, so the plain $n!$ overcounts. If among $n$ objects one kind repeats $p$ times, another $q$ times, another $s$ times, the distinct arrangements number:
$$\dfrac{n!}{p!\,q!\,s!}$$
Restrictions are the heart of exam permutation problems. The three classic types and their reliable strategies are:
- Objects together ("vowels together", "two people side by side"): glue the bound objects into a single block, arrange the block alongside the free objects, then multiply by the internal arrangements of the block.
- Objects never together (separation): first arrange the other objects, which creates gaps between and around them, then place the objects to be separated into distinct gaps — this guarantees no two are adjacent. Alternatively, use (total) $-$ (together).
- Fixed positions ("begins with a consonant", "ends in a vowel"): fill the restricted positions first with their allowed objects, then fill the remaining positions with what is left.
Deeper Insight — every permutation rule is really a correction to $n!$: It helps to see all the formulas as one idea under different conditions. The starting point is $n!$, the count of arrangements when every object is distinct and all are used. From there you adjust. If you only fill $r$ of the slots, you divide away the arrangements of the $n-r$ leftovers, giving $\dfrac{n!}{(n-r)!}$. If objects may repeat, you abandon the descending product entirely and use $n^r$, because nothing is ever "used up". If some objects are identical, you divide out the $p!,\,q!,\dots$ ways those clones could swap among themselves without changing what you see. Recognising which adjustment a problem demands — and resisting the urge to plug into the first formula you remember — is the skill that separates a confident solver from a guesser. The single question that settles it every time: does order matter, and are the objects truly distinct?
In how many ways can the offices of President, Secretary, and Treasurer be filled from a club of 10 members (no person holds two offices)?
Solution- The three offices are distinct, so order matters — this is a permutation of $10$ taken $3$ at a time.
- ${}^{10}P_{3} = \dfrac{10!}{(10-3)!} = \dfrac{10!}{7!} = 10 \times 9 \times 8$.
- $10 \times 9 = 90$, and $90 \times 8 = 720$.
Answer: $720$ ways.
How many distinct arrangements can be made of the letters of the word MATHEMATICS?
Solution- MATHEMATICS has $11$ letters. Count repeats: M appears $2$ times, A appears $2$ times, T appears $2$ times; H, E, I, C, S appear once each.
- With objects not all distinct, the count is $\dfrac{11!}{2!\,2!\,2!}$.
- $11! = 39{,}916{,}800$ and $2!\,2!\,2! = 8$.
- $\dfrac{39{,}916{,}800}{8} = 4{,}989{,}600$.
Answer: $4{,}989{,}600$ distinct arrangements.
How many 4-letter codes can be formed from the 26 letters of the alphabet if letters may be repeated?
Solution- Repetition is allowed, so each of the $4$ positions independently has all $26$ letters available.
- By the multiplication principle the count is $26 \times 26 \times 26 \times 26 = 26^{4}$.
- $26^2 = 676$, and $676^2 = 456{,}976$.
Answer: $26^{4} = 456{,}976$ codes.
In how many ways can the letters of the word EQUATION be arranged so that all the vowels come together?
Solution- EQUATION has $8$ distinct letters. The vowels are E, U, A, I, O (five vowels); the consonants are Q, T, N (three consonants).
- Glue the five vowels into a single block. We now arrange this block together with the $3$ consonants — that is $4$ items in a row: $4! = 24$ ways.
- Within the block, the $5$ vowels can be ordered among themselves in $5! = 120$ ways.
- Multiply: $4! \times 5! = 24 \times 120 = 2{,}880$.
Answer: $2{,}880$ arrangements.
Find $n$ if ${}^{n}P_{4} = 20 \times {}^{n}P_{2}$.
Solution- Write each side out: ${}^{n}P_{4} = n(n-1)(n-2)(n-3)$ and ${}^{n}P_{2} = n(n-1)$.
- The equation becomes $n(n-1)(n-2)(n-3) = 20\,n(n-1)$.
- Since $n \ge 4$, divide both sides by $n(n-1)$: $(n-2)(n-3) = 20$.
- Expand: $n^2 - 5n + 6 = 20 \Rightarrow n^2 - 5n - 14 = 0 \Rightarrow (n-7)(n+2) = 0$.
- So $n = 7$ or $n = -2$; reject the negative root.
Answer: $n = 7$.
How many 5-digit numbers can be formed from the digits $1, 2, 3, 4, 5, 6, 7$ without repetition such that the number is even?
Solution- For the number to be even, the units digit must be one of $2, 4, 6$ — that is $3$ choices for the last position.
- The remaining $4$ positions are filled from the $6$ digits left, in order, without repetition: ${}^{6}P_{4} = 6 \times 5 \times 4 \times 3 = 360$.
- Multiply by the choices for the units digit: $360 \times 3 = 1{,}080$.
Answer: $1{,}080$ even numbers.
In how many ways can $7$ people be seated in a row so that two particular people, A and B, are never seated next to each other?
Solution- Count the total arrangements, then subtract those in which A and B are together — the complement is easier here.
- Total arrangements of $7$ people: $7! = 5{,}040$.
- Together: glue A and B into a block, giving $6$ items to arrange in $6! = 720$ ways, times $2!$ for A–B order inside the block: $720 \times 2 = 1{,}440$.
- Never together $= 5{,}040 - 1{,}440 = 3{,}600$.
Answer: $3{,}600$ arrangements.
How many words can be formed from the letters of ARRANGE? How many of these have the two R's never together?
Solution- ARRANGE has $7$ letters with A repeated $2$ times and R repeated $2$ times. Total distinct arrangements $= \dfrac{7!}{2!\,2!} = \dfrac{5040}{4} = 1{,}260$.
- Arrangements with the two R's together: glue them into one block (the two R's are identical, so no internal $2!$). We now arrange $6$ items, of which A repeats twice: $\dfrac{6!}{2!} = \dfrac{720}{2} = 360$.
- R's never together $= 1{,}260 - 360 = 900$.
Answer: $1{,}260$ words in total; $900$ have the two R's never together.
In how many ways can the letters of the word PERMUTATIONS be arranged if the word must begin with P and end with S?
Solution- PERMUTATIONS has $12$ letters; the only repeated letter is T (appears $2$ times).
- Fix P in the first position and S in the last — both are now used and settled.
- The remaining $10$ letters (E, R, M, U, T, A, T, I, O, N, with T repeated twice) fill the $10$ middle positions: $\dfrac{10!}{2!}$.
- $\dfrac{3{,}628{,}800}{2} = 1{,}814{,}400$.
Answer: $1{,}814{,}400$ arrangements.
In how many ways can $4$ boys and $3$ girls be seated in a row so that no two girls sit together?
Solution- Seat the $4$ boys first: they can be arranged in $4! = 24$ ways.
- Four boys create $5$ gaps (including the two ends) where girls may go: $\_\,B\,\_\,B\,\_\,B\,\_\,B\,\_$.
- Place the $3$ girls into $3$ of these $5$ gaps, order mattering, with at most one girl per gap: ${}^{5}P_{3} = 5 \times 4 \times 3 = 60$.
- Multiply: $24 \times 60 = 1{,}440$.
Answer: $1{,}440$ arrangements.
How many $4$-digit numbers can be formed using the digits $0,1,2,3,4,5$ if repetition of digits is allowed? (No leading zero.)
Solution- Repetition is allowed, so digits are not "used up", but the thousands place still cannot be $0$.
- Thousands place: $5$ choices ($1$–$5$).
- Each of the other three places: all $6$ digits are available, so $6$ choices each.
- Total $= 5 \times 6 \times 6 \times 6 = 5 \times 216 = 1{,}080$.
Answer: $1{,}080$ four-digit numbers.
Find $r$ if ${}^{5}P_{r} = 2 \times {}^{6}P_{r-1}$.
Solution- Write each in factorial form: $\dfrac{5!}{(5-r)!} = 2 \times \dfrac{6!}{(6-(r-1))!} = 2 \times \dfrac{6!}{(7-r)!}$.
- Substitute $6! = 6 \times 5!$ and cancel $5!$: $\dfrac{1}{(5-r)!} = \dfrac{12}{(7-r)!}$.
- Note $(7-r)! = (7-r)(6-r)(5-r)!$, so cancelling $(5-r)!$ gives $(7-r)(6-r) = 12$.
- Let $k = 6-r$: $(k+1)k = 12 \Rightarrow k^2 + k - 12 = 0 \Rightarrow (k+4)(k-3)=0$, so $k = 3$ (taking the valid root), giving $r = 3$.
Answer: $r = 3$.