Counting principle and permutations

Suppose a canteen offers 4 starters, 5 mains, and 3 desserts, and you want to know how many complete meals someone could choose — or how many distinct ways 10 people can line up for a photo. Listing every possibility by hand would take forever, and it is exactly the kind of problem where a small set of counting rules replaces brute-force listing with a handful of multiplications. This is the promise of the fundamental principle of counting: break a complicated choice into independent stages, count the options at each stage, and multiply. From that one idea, everything else in this unit follows — factorial notation as shorthand for "arrange everything," the permutation formula nPr^nP_r for arranging a subset of objects in order, and correction factors for when repetition is allowed or when some objects are identical (like the repeated letters in a word). This topic opens the Permutations and Combinations sub-unit of Algebra in ISC Class 11 Mathematics, and it is the launchpad for everything that follows — restricted arrangements, circular permutations, and combinations. Master the multiplication principle and the nPr^nP_r formula here, and the rest of the unit becomes pattern-matching rather than new theory.

The core insight

If a task splits into independent stages with mm, nn, ... choices each, the total number of ways is the product m×n×⋯m\times n\times\cdots, and this single fact generates factorials, the permutation formula nPr=n!/(n−r)!^nP_r=n!/(n-r)!, and their variants for repetition and identical objects.

The core idea

Multiplication versus addition. Two tasks done one after another, independently, combine by multiplying the choice counts; two mutually exclusive alternatives (only one of which happens) combine by adding them. The single word that separates the two rules in a word problem is usually "and" (multiply) versus "or" (add).

Factorial notation. Arranging nn distinct objects in a row uses every object exactly once, and the count of such arrangements is written n!n!, read "nn factorial":

n!=n(n−1)(n−2)⋯2⋅1,0!=1 (by convention).n! = n(n-1)(n-2)\cdots 2\cdot 1,\qquad 0! = 1 \text{ (by convention)}.

where: nn is a non-negative integer; the convention 0!=10!=1 keeps later formulas consistent when zero objects remain to be arranged.

Permutations of a subset. Often only rr of the nn available objects are placed, and order still matters (a permutation, as opposed to a combination, cares about order — first place differs from second place). Applying the multiplication principle stage by stage — nn choices for the first position, n−1n-1 for the second, and so on down to n−r+1n-r+1 choices for the rr-th — gives the count directly, and dividing out the unused tail of the factorial gives the closed form used in the derivation below.

Two important variants. When repetition of objects is allowed, each of the rr positions independently has all nn choices available, giving nrn^r arrangements — no factorial needed. When some of the nn objects being arranged are identical (repeated letters in a word, say), swapping identical objects among themselves does not create a visibly new arrangement, so the naive count n!n! must be divided by p! q!⋯p!\,q!\cdots, where p,q,…p,q,\ldots are the multiplicities of each repeated kind.

Figure
Three counting scenarios shown side by side: independent stages multiplying to a total, a permutation formula reducing a factorial, and identical objects shrinking the count by dividing out repeated-symbol factorials
Three faces of the counting principle independent stages 4 x 5 x 3 = 60 meals permutation nPr n! / (n-r)! 8P3 = 336 identical objects n! / (p! q! ...) STATISTICS: 50400 All three rules trace back to the same multiplication principle applied stage by stage.
Derivation ADeriving the permutation formula nPr

Derive the closed-form expression for the number of ways to arrange rr objects chosen, without repetition, from a pool of nn distinct objects when the order of placement matters, by directly applying the fundamental principle of counting stage by stage and then rewriting the resulting product using factorial notation.

Step 1. Picture rr positions in a row to be filled, one object per position, all objects drawn from the same pool of nn distinct objects with no object reused. Filling the first position is a single independent choice with nn options, since every object in the pool is still available.

Step 2. Once the first position is filled, one object has been used up, so the second position has only n−1n-1 objects left to choose from. By the multiplication principle, the number of ways to fill the first two positions together is n(n−1)n(n-1), since the two choices are made in independent successive stages.

Step 3. Continuing this pattern, the third position has n−2n-2 objects left, the fourth has n−3n-3, and in general the kk-th position has n−(k−1)n-(k-1) objects available, because k−1k-1 objects have already been placed and removed from the pool. Multiplying the choice counts across all rr positions gives

nPr=n(n−1)(n−2)⋯(n−r+1),^{n}P_r = n(n-1)(n-2)\cdots(n-r+1),

a product of exactly rr consecutive descending factors starting at nn.

Step 4. This product can be written more compactly using factorial notation. Multiply and divide by the "missing tail" (n−r)(n−r−1)⋯2⋅1=(n−r)!(n-r)(n-r-1)\cdots2\cdot1 = (n-r)!, which does not appear in the product above:

nPr=n(n−1)(n−2)⋯(n−r+1)⋅(n−r)!(n−r)!=n!(n−r)!.^{n}P_r = \frac{n(n-1)(n-2)\cdots(n-r+1)\cdot(n-r)!}{(n-r)!} = \frac{n!}{(n-r)!}.

The numerator is now the full product of every integer from nn down to 11, which is exactly n!n!, so the formula is confirmed:

nPr=n!(n−r)!,0≤r≤n.^{n}P_r = \frac{n!}{(n-r)!}, \qquad 0\le r\le n.

As a check, setting r=nr=n gives nPn=n!/0!=n!^nP_n = n!/0! = n! (using 0!=10!=1), matching the direct count of arranging all nn objects in Step 3, and setting r=0r=0 gives nP0=n!/n!=1^nP_0 = n!/n! = 1, matching the single "do nothing" arrangement.

Worked example

A café offers 4 starters, 5 mains, and 3 desserts, and separately, 6 people are to be seated in a row of 6 distinct chairs. Find both counts, then use the same reasoning to find the number of distinct arrangements of the letters of STATISTICS and the value of 8P3^8P_3.

Step 1 (café choices). The three course choices are independent successive tasks — pick a starter, and a main, and a dessert — so by the fundamental principle of counting the total number of complete meals is 4×5×3=604\times5\times3 = 60.

Step 2 (seating 6 people). Seating all 6 people in 6 distinct chairs is a permutation of all 6 objects, so 6P6=6!=6×5×4×3×2×1=720^6P_6 = 6! = 6\times5\times4\times3\times2\times1 = 720.

Step 3 (letters of STATISTICS). The word has 10 letters total, with S repeated 3 times, T repeated 3 times, I repeated 2 times, and A and C appearing once each. Using the identical-objects formula, the number of distinct arrangements is

10!3! 3! 2! 1! 1!=36288006×6×2=362880072=50400.\frac{10!}{3!\,3!\,2!\,1!\,1!} = \frac{3628800}{6\times6\times2} = \frac{3628800}{72} = 50400.

Step 4 (8P3^8P_3 and a sanity check). Directly from the derived formula, 8P3=8!5!=8×7×6=336^8P_3 = \dfrac{8!}{5!} = 8\times7\times6 = 336. As a cross-check on Step 3: if all 10 letters of STATISTICS were distinct there would be 10!=362880010! = 3628800 arrangements, but the repeats overcount every genuine arrangement by a factor of 3!×3!×2!=6×6×2=723!\times3!\times2! = 6\times6\times2 = 72, and 3628800/72=504003628800/72 = 50400 again — confirming the answer.

Final answers: 60 meals, 6P6=720^6P_6 = 720 seatings, 50400 distinct arrangements of STATISTICS, and 8P3=336^8P_3 = 336.

Competitive stretch

A quicker way to remember Step 4's cross-check: for any word, distinct arrangements always equal (all-distinct count n!n!) divided by (product of repeat-factorials), so a one-line sanity pass — count repeats, multiply their factorials, divide into n!n! — catches most arithmetic slips before they become wrong final answers.

Common traps

  • Confusing "and" (independent successive choices — multiply) with "or" (mutually exclusive alternatives — add); misreading the connective in a word problem is the single most common error in this topic.
  • Using nrn^r (repetition allowed) when the problem forbids repeated objects, or using nPr^nP_r when repetition is actually allowed — always check the repetition condition before picking a formula.
  • Forgetting the identical-objects correction factor p! q!⋯p!\,q!\cdots for repeated letters or items, which overcounts every arrangement that differs only by swapping identical objects.
  • Miscomputing 0!0! as 00 instead of 11 — this breaks the nPr^nP_r formula whenever r=nr=n, since (n−n)!=0!=1(n-n)! = 0! = 1 by convention, not 00.

Ask the tutor: If a word problem does not clearly say whether repetition is allowed or whether objects repeat, ask the AI tutor to help you spot the clue phrase before you commit to a formula.

Why does dividing by (n minus r) factorial in the nPr formula actually work

Writing out nPr^nP_r as the product n(n−1)⋯(n−r+1)n(n-1)\cdots(n-r+1) shows it stops exactly rr factors after nn. The full factorial n!n! continues all the way down to 11, so it has an extra "tail" of (n−r)(n-r) factors that nPr^nP_r does not want. Dividing n!n! by (n−r)!(n-r)! cancels that unwanted tail exactly, leaving only the rr factors you actually multiplied in the stage-by-stage count — that is why the two expressions agree.

How do I quickly tell whether a problem wants nPr, n to the r, or the identical-objects formula

Ask two questions. First, can any object be reused across positions? If yes, it is repetition-allowed and the count is nrn^r; if no, move to the second question. Second, are the nn objects themselves all distinct, or are some identical? If all distinct, use nPr=n!/(n−r)!^nP_r = n!/(n-r)!; if some repeat (like letters in a word), arrange all of them and divide by the factorial of each repeat count, n!/(p! q!⋯ )n!/(p!\,q!\cdots). Practising this two-question check on a handful of past-paper problems makes the choice automatic.

Is a permutation the same thing as a combination

No — order is the deciding factor. A permutation counts arrangements where the sequence matters, so choosing A then B is different from choosing B then A (like assigning first and second prize). A combination only cares which objects are chosen, not their order, so AB and BA count as the same selection (like picking a 2-person committee). Because every combination can be arranged in more than one order, nPr^nP_r is always at least as large as the corresponding combination count nCr^nC_r, in fact nPr=nCr×r!^nP_r = {^nC_r}\times r!.

Before this

Where this leads