Permutations and Combinations

This chapter climbs from one counting principle to two big formulas. It opens with the fundamental principle of counting — the multiplication principle, illustrated with pants-and-shirts and coloured-flag problems — and its natural companion, the addition principle for mutually exclusive cases. It then defines the permutation, introduces factorial notation, derives ${}^{n}P_r = n!/(n-r)!$ for arrangements of distinct objects, and extends this to arrangements with repetition ($n^r$) and to arrangements of objects that are not all distinct (words with repeated letters, discs of the same colour, digits that recur). It then turns to the combination, defines ${}^{n}C_r$, proves the link ${}^{n}P_r = {}^{n}C_r \, r!$, records the symmetry ${}^{n}C_r = {}^{n}C_{n-r}$, and proves Pascal's rule ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$. Worked examples throughout mirror NCERT's own numbered examples on lock combinations, word arrangements, restricted seating, committee selections and card-hand counting — the chapter's stated goal is always to count without listing.

The fundamental principle of counting

Suppose Mohan owns 3 pants and 2 shirts. In how many different ways can he dress in one pant and one shirt? A pant can be chosen in 3 ways; whichever pant is chosen, a shirt can then be chosen in 2 ways; so the pair can be chosen in $3 \times 2 = 6$ ways. This is the seed of the whole chapter, generalised to any finite chain of successive events.

where: $m_i$ is the number of ways the $i$-th event can occur, counted after every earlier event has already occurred, provided that number of available ways stays the same no matter which of the earlier choices was actually made. Filling vacant places in succession is the standard picture: the first place can be filled in $m_1$ ways, and for each of those, the second place can be filled in $m_2$ ways, and so on, so the total number of ways of filling all the places is the product. A companion rule, the addition principle, handles jobs that can be done in one of several mutually exclusive ways: if a task can be completed by method one in $a$ ways or by an entirely separate method two in $b$ ways, with no overlap, the total is $a+b$. NCERT's own flag-signal problem uses exactly this: with 5 different flags, a signal of 2 flags can be made in $5 \times 4 = 20$ ways, of 3 flags in $5\times4\times3=60$ ways, of 4 flags in $120$ ways and of 5 flags in $120$ ways, so a signal of at least 2 flags — an "either/or" of four disjoint cases — totals $20+60+120+120=320$ by addition.

Permutations of distinct objects: factorial notation and nPr

An arrangement of objects in a definite order is a permutation. If four different letters, R, O, S, E, are arranged in all possible orders taking all four at a time, each distinct ordering (ROSE, REOS, OSER, ...) counts as a separate permutation, because order matters: ROSE and OSER use the same letters but are different arrangements. Counting how many words of 3 letters can be formed from the 6 different letters of NUMBER, without repeating a letter, is asking for the permutations of 6 distinct objects taken 3 at a time, and the multiplication principle gives $6 \times 5 \times 4 = 120$ directly, by filling three vacant places in succession. This repeated pattern of "filling $r$ vacant places from $n$ available objects, without repetition" is compressed into a single symbol and a single formula.

where: $n$ is the number of distinct objects available, $r$ is the number chosen and arranged in order, and $n!$ ("$n$ factorial") is the product of the first $n$ natural numbers, with $0! = 1$ fixed by convention so that equation $(3)$ still gives the sensible value ${}^{n}P_0 = 1$ (there is exactly one way to arrange nothing at all — leave every object where it is). The full proof that the messy product $n(n-1)(n-2)\cdots(n-r+1)$ collapses into the tidy ratio $n!/(n-r)!$ is given in the first derivation below. When repetition of the $n$ objects is allowed — each of the $r$ positions can independently be any of the $n$ objects, with no restriction against reuse — the count changes entirely, because every position again has a full $n$ choices regardless of what filled the earlier positions.

where: each of the $r$ positions is filled independently from the same pool of $n$ objects, so the $r$ factors of $n$ multiply together $r$ times. For the word ROSE with repetition allowed, the count of 4-letter words rises from ${}^{4}P_4 = 24$ (no repetition) to $4^4 = 256$ (repetition allowed) — a useful sanity check that repetition always increases (or, at worst, matches) the no-repetition count.

Permutations when the objects are not all distinct

Arranging the letters of ROOT is different from arranging four fully distinct letters, because the two O's are identical: swapping them produces the same written word, not a new one. Treat the two O's as temporarily distinguishable ($O_1, O_2$) to count $4!$ arrangements, then notice every genuine word (such as ROOT) has been counted exactly $2!$ times — once for each way of labelling the identical O's — so the true count is $4!/2! = 12$. Extending this reasoning to several repeated kinds of objects at once gives the general formula.

where: $n$ is the total number of objects being arranged, and $p_1, p_2, \ldots, p_k$ are the multiplicities of each repeated kind (objects of a kind that appears only once contribute a factor of $1! = 1$ and can be ignored). For the word ALLAHABAD — 9 letters with 4 A's and 2 L's, the rest distinct — the number of arrangements is $9!/(4!\,2!) = 7560$. This same formula underlies restricted permutations, such as requiring a set of letters to occur together: bundle the restricted letters into one block, treat the block plus the remaining letters as a smaller set of objects to arrange (accounting for any repeats among them), and then multiply by the number of ways to arrange the contents inside the block — a two-stage multiplication-principle argument fully worked through in the second worked example below.

Combinations: selections where order does not matter

Three lawn-tennis players X, Y, Z are to be reduced to a team of 2. Is the team "X and Y" different from the team "Y and X"? No — a team is just a set of people, with no order attached, so there are only 3 possible teams: XY, YZ, ZX. Each such selection is a combination of 3 objects taken 2 at a time, and the essential difference from a permutation is precisely this: a combination does not care which object was "chosen first."

where: $n$ is the pool size, $r$ is how many are selected with no regard to order, and dividing ${}^{n}P_r$ by $r!$ removes the overcounting caused by the $r!$ internal orderings that each unordered selection would otherwise be split across — proved carefully in the second derivation below. Three immediate consequences follow directly from equation $(6)$: ${}^{n}C_0 = {}^{n}C_n = 1$ (there is exactly one way to select nothing, and exactly one way to select everything); the symmetry ${}^{n}C_r = {}^{n}C_{n-r}$, because choosing $r$ objects to keep is the same act as choosing $n-r$ objects to reject; and if ${}^{n}C_a = {}^{n}C_b$ with $a \neq b$, then necessarily $a + b = n$. Combinations also satisfy an addition identity linking one row of counts to the next, Pascal's rule, proved in full in the third derivation below.

where: the left side adds two consecutive combination counts for a pool of size $n$, and the right side is a single combination count for a pool one object larger — this identity is the algebraic seed of Pascal's triangle and reappears as the key step in proving the binomial theorem in the next chapter.

Board level - solving for n from a permutation equation (Example 12 pattern)

Find the value of $n$ such that ${}^{n}P_5 = 42\, {}^{n}P_3$, given $n > 4$.

Step 1. Write ${}^{n}P_5$ as ${}^{n}P_3$ multiplied by two extra descending factors, since filling 5 places is the same as filling 3 places and then 2 more: ${}^{n}P_5 = n(n-1)(n-2)(n-3)(n-4)$ and ${}^{n}P_3 = n(n-1)(n-2)$, so the equation becomes $n(n-1)(n-2)(n-3)(n-4) = 42\,n(n-1)(n-2)$.

Step 2. Because $n>4$, the factors $n$, $(n-1)$ and $(n-2)$ are all non-zero, so they can be safely cancelled from both sides, leaving the much smaller equation $(n-3)(n-4) = 42$.

Step 3. Expand and rearrange into standard quadratic form: $n^2 - 7n + 12 = 42$, so $n^2 - 7n - 30 = 0$. Factorise: $(n-10)(n+3) = 0$, giving $n = 10$ or $n = -3$.

Step 4. Reject $n=-3$, since $n$ counts objects and cannot be negative (and the problem states $n>4$ in any case). Answer: n = 10.

Board level - vowels together and vowels apart in a word with distinct letters (Example 14 pattern)

Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together, and (ii) all vowels do not occur together.

Step 1. DAUGHTER has 8 distinct letters, of which 3 are vowels (A, U, E) and 5 are consonants (D, G, H, T, R). For part (i), treat the block of 3 vowels as glued into a single unit, temporarily written AUE. This single block, together with the 5 separate consonants, gives 6 objects in total to arrange.

Step 2. Arrange these 6 objects (5 consonants plus 1 vowel-block) in a row: since all 6 are distinct objects at this stage, this is ${}^{6}P_6 = 6! = 720$ ways. But the vowel-block itself is not a rigid unit — its 3 internal letters can be reordered among themselves in ${}^{3}P_3 = 3! = 6$ ways for every one of those 720 outer arrangements.

Step 3. By the multiplication principle, the number of arrangements with all vowels together is $6! \times 3! = 720 \times 6 = 4320$. For part (ii), first find the total number of arrangements with no restriction at all: since all 8 letters of DAUGHTER are distinct, this is $8! = 40320$. The arrangements where vowels do not all occur together are everything except the ones where they do.

Step 4. Subtract: $40320 - 4320 = 36000$. Answer: vowels together gives 4320 arrangements; vowels not all together gives 36000 arrangements.

Competitive level (JEE / NEET) - team selection with an at least one of each restriction, via complementary counting (Example 21 pattern)

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected so that the team contains at least one boy and at least one girl?

Step 1. The direct route is to split into the four disjoint cases (1 boy and 4 girls), (2 boys and 3 girls), (3 boys and 2 girls), (4 boys and 1 girl), and add ${}^{7}C_1{}^{4}C_4 + {}^{7}C_2{}^{4}C_3 + {}^{7}C_3{}^{4}C_2 + {}^{7}C_4{}^{4}C_1$. That works, but it is four separate combination products to compute under time pressure. The faster competitive-exam route is complementary counting: count everything, then subtract only the selections that fail the restriction.

Step 2. The pool has $4+7=11$ people in total, so the number of ways to pick any 5 of them, with no restriction at all, is ${}^{11}C_5 = \dfrac{11!}{5!\,6!} = \dfrac{11\times10\times9\times8\times7}{120} = 462$.

Step 3. A team fails "at least one boy and one girl" in exactly two mutually exclusive ways: it has no girls at all (all 5 boys), or no boys at all (all 5 girls). All-boys teams number ${}^{7}C_5 = 21$. All-girls teams are impossible, since only 4 girls exist and a team of 5 cannot be built from them: ${}^{4}C_5 = 0$.

Step 4. Subtract both failing cases from the unrestricted total: $462 - 21 - 0 = 441$. As a cross-check, the four-case direct sum from Step 1 gives ${}^{7}C_1{}^{4}C_4 + {}^{7}C_2{}^{4}C_3 + {}^{7}C_3{}^{4}C_2 + {}^{7}C_4{}^{4}C_1 = 7 + 84 + 210 + 140 = 441$, exactly matching. Answer: 441 ways. Timing note: budget about ninety seconds. Reach for complementary counting the instant you see "at least one of each" across two categories — it collapses four combination products into one subtraction. Always check whether a "failing" case is impossible (as ${}^{4}C_5=0$ is here) before spending time computing it.

The single most frequent error is applying ${}^{n}P_r$ to a problem that is actually asking for a selection, or vice versa — ask first whether swapping the order of the chosen objects gives a genuinely different outcome (shelf: yes; committee: no). A second common slip appears with repeated-letter words: students correctly compute $n!$ but forget to divide by the factorial of each repeated letter's multiplicity, silently overcounting identical-looking arrangements. A third slip is with restricted arrangements such as "letters together": students bundle the group into one block and arrange the blocks, but forget the second multiplication step — permuting the contents inside the block. In "at least/at most" problems, a frequent error is missing a boundary case (forgetting "at least 3" includes "exactly 4" when only 4 are available) or double-counting overlapping cases instead of splitting them into mutually exclusive groups before adding. Finally, students often mishandle $0!$, treating it as $0$ instead of the defined value $1$, which silently breaks the boundary cases ${}^{n}P_0$ and ${}^{n}P_n$.

• State the fundamental principle of counting precisely, and apply it to fill vacant positions in succession for problems with and without repetition.
• Define $n!$, know that $0!=1$ by convention, and simplify factorial expressions such as $12!/(10!\times 2!)$ without expanding every factor.
• State the formula ${}^{n}P_r = n!/(n-r)!$, derive it from the multiplication principle, and use it to count arrangements of distinct objects.
• Distinguish permutations with repetition ($n^r$) from permutations without repetition (${}^{n}P_r$), and know which applies to a given word problem.
• Count arrangements of objects that are not all distinct using $n!/(p_1!\,p_2!\cdots p_k!)$, correctly identifying every repeated letter or object and its multiplicity.
• Solve restricted-arrangement problems (letters or people kept together, or kept apart) by the block-and-multiply technique, including the internal-arrangement step.
• State the formula ${}^{n}C_r = n!/(r!(n-r)!)$, derive it from ${}^{n}P_r = {}^{n}C_r\, r!$, and correctly decide, for any given problem, whether it needs a permutation or a combination.
• Use the symmetry ${}^{n}C_r = {}^{n}C_{n-r}$ and Pascal's rule ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$, and prove Pascal's rule algebraically from the factorial definition.
• Solve committee and card-hand problems with restrictions ("at least," "at most," "exactly") by splitting into mutually exclusive cases or by complementary counting.
• Solve algebraic equations in $n$ or $r$ built from ${}^{n}P_r$ or ${}^{n}C_r$ by cancelling common non-zero factors before expanding.

Ask the tutor: Want to work through a trickier restricted-arrangement or at-least-committee problem together, step by step?