Permutations of distinct objects

A class election needs a president, a secretary and a treasurer chosen from 12 students, no one holding two posts. How many different outcomes are possible? This is the cleanest question permutations answer: given nn objects that are all distinct from one another, how many ordered arrangements can you make by choosing rr of them at a time, each used at most once? It sits inside the CBSE Class 11 chapter Permutations and Combinations as the foundational case — no repeated objects, no reused positions — from which the repetition-allowed and not-all-distinct cases are later built by adjustment. This article stays inside that one case. We derive the formula nPr=n!(n−r)!^nP_r = \dfrac{n!}{(n-r)!} from the fundamental principle of counting, work through the office-bearer and bookshelf examples in full, and flag the traps that cost the most marks: confusing this with repetition, and confusing ordered posts with unordered groups.

The core insight

For nn distinct objects with none repeated and none reused, the number of ordered arrangements of rr of them is nPr=n!(n−r)!^nP_r = \dfrac{n!}{(n-r)!}, built by multiplying nn shrinking choices for rr positions in a row.

The core idea

The setup. Take nn objects that are all different from each other — 12 distinct students, 5 distinct books, the 10 distinct digits. Choose rr of them (with r≤nr \le n) and arrange them in a row, each object used exactly once in that arrangement. Two arrangements count as different if the order differs, even with the same objects — this is what separates a permutation from a combination.

The counting argument. Fill rr positions one at a time using the multiplication principle. Position 1 can take any of the nn objects. Position 2 can take any of the remaining n−1n-1. Position 3 takes any of the remaining n−2n-2, and so on down to position rr, which has n−r+1n-r+1 choices left. Multiplying these shrinking counts together gives the arrangement count directly, and dividing and multiplying by (n−r)!(n-r)! turns the product into a clean factorial ratio.

nPr=n!(n−r)!.^nP_r = \frac{n!}{(n-r)!}.

where: nn is the total number of distinct objects available, rr is how many are chosen and arranged (positions filled), and n!n! (n factorial) is n(n−1)(n−2)⋯1n(n-1)(n-2)\cdots1, with the convention 0!=10! = 1.

Two special cases. Setting r=nr = n (arrange all the objects) gives nPn=n!^nP_n = n!, matching the direct intuition of nn choices, then n−1n-1, down to 11. Setting r=0r = 0 gives nP0=1^nP_0 = 1, meaning there is exactly one way to arrange nothing at all.

Why the formula only fits this case. Because every object here is genuinely distinct and each one is used at most once, there is no double-counting to correct for anywhere in the product. That is precisely what makes nPr=n!(n−r)!^nP_r = \dfrac{n!}{(n-r)!} the simplest of the three permutation formulas in the chapter; the versions for repetition allowed and for objects that are not all distinct start from this same shrinking-choices idea and then adjust it, either by not shrinking the choices (repetition) or by dividing out arrangements that look identical (repeated objects).

Figure
Filling three positions from twelve distinct students the choices shrink from twelve to eleven to ten
Filling 3 posts from 12 distinct students President 12 choices Secretary 11 choices Treasurer 10 choices 12 x 11 x 10 = 1320 arrangements Each chosen student is removed from later choices — nobody repeats
Derivation Aof the permutation formula

The goal is to show, using only the fundamental principle of counting (the "multiply the choices" rule already familiar from earlier in the chapter), that arranging rr objects chosen from nn distinct objects, one per position and no repeats, must equal the product n(n−1)(n−2)⋯(n−r+1)n(n-1)(n-2)\cdots(n-r+1), and that this product simplifies exactly to n!(n−r)!\dfrac{n!}{(n-r)!} once the missing lower factorial terms are supplied.

Step 1. Set up rr positions in a row, to be filled left to right, and count the choices available at each one. The first position can be filled by any of the nn distinct objects, so there are nn ways to fill it.

Step 2. Once position 1 is filled, that object is no longer available (no repetition), so position 2 has only n−1n-1 objects left to choose from. Position 3 then has n−2n-2 objects left, since two objects are now used. This pattern continues: at position kk, exactly k−1k-1 objects have already been placed, leaving n−(k−1)=n−k+1n-(k-1) = n-k+1 choices.

Step 3. Continue until the last, rr-th position, which is filled from whatever remains after r−1r-1 objects have already been placed: n−(r−1)=n−r+1n - (r-1) = n - r + 1 choices. By the multiplication principle, the total number of ways to fill all rr positions is the product of the choices at each step:

nPr=n(n−1)(n−2)⋯(n−r+1).^nP_r = n(n-1)(n-2)\cdots(n-r+1).

Step 4. Convert this product into factorial form by multiplying and dividing by (n−r)!(n-r)!, which is exactly the product of the remaining terms from (n−r)(n-r) down to 11 that the product above stops short of:

nPr=n(n−1)⋯(n−r+1) (n−r)!(n−r)!=n!(n−r)!.^nP_r = \frac{n(n-1)\cdots(n-r+1)\,(n-r)!}{(n-r)!} = \frac{n!}{(n-r)!}.

The numerator's expansion collapses to n!n! because it now runs unbroken from nn down to 11, giving the compact closed form used throughout the chapter.

Worked example

Problem. In how many ways can a president, a secretary and a treasurer be chosen from a class of 12 distinct students, if no student can hold more than one post? Extend this: five distinct books A, B, C, D, E are to be arranged in a row on a shelf — in how many ways can all five be arranged, and in how many of those arrangements is book A placed first?

Step 1. Identify the election as a permutation problem: three different roles are being filled from 12 distinct students, order matters (president is not the same as secretary), and no student repeats. So n=12n = 12, r=3r = 3.

Step 2. Apply the formula: 12P3=12!9!=12×11×10=1320^{12}P_3 = \dfrac{12!}{9!} = 12 \times 11 \times 10 = 1320. So there are 1,320 ways to fill the three posts.

Step 3. For the bookshelf, arranging all five distinct books uses the special case r=n=5r = n = 5: 5P5=5!=5×4×3×2×1=120^5P_5 = 5! = 5\times4\times3\times2\times1 = 120 total orders.

Step 4. With book A fixed in the first position, only the remaining 4 distinct books fill the remaining 4 positions: 4P4=4!=24^4P_4 = 4! = 24. Answer: 1,320 ways to fill the three posts; 120 total shelf orders, of which 24 have book A first — exactly one-fifth, matching the symmetry that A is equally likely to land in any of the 5 spots.

Competitive stretch

How many 4-digit codes can be formed from the digits 11 to 99 (no repetition, no leading-zero issue since 00 is excluded)? This is 9P4=9×8×7×6=3024^9P_4 = 9\times8\times7\times6 = 3024 codes — the same shrinking-choices logic, just with digits instead of people.

Common traps

  • Using n!(n−r)!\dfrac{n!}{(n-r)!} when some objects actually repeat (like repeated letters in a word) — that over-counts; the not-all-distinct case needs n!p1!p2!⋯pk!\dfrac{n!}{p_1!p_2!\cdots p_k!} instead.
  • Confusing this with the repetition-allowed case, where reuse is permitted and the count is nrn^r, not n!(n−r)!\dfrac{n!}{(n-r)!}.
  • Treating distinguishable roles or positions (president/secretary/treasurer, 1st/2nd/3rd place) as an unordered group — that mistake turns a permutation problem into a (wrong) combination count.
  • Arithmetic slips computing (n−r)!(n-r)! or cancelling factorials for larger nn; it is safer to expand nPr^nP_r directly as n(n−1)⋯(n−r+1)n(n-1)\cdots(n-r+1) than to compute two large factorials separately.

Ask the tutor: want to try a mixed set of problems that force you to pick between this formula, the repetition-allowed case, and combinations?

Why do we divide by (n-r)! instead of just multiplying n times itself r times?

Multiplying n by itself r times, nrn^r, would count arrangements where the same object is reused in more than one position — that is the repetition-allowed case, not this one. Here every object is used at most once, so the choices shrink at each step: n, then n-1, then n-2, and so on. Dividing n! by (n-r)! is just a compact way of writing that shrinking product, since (n-r)! is exactly the leftover tail of the factorial that the shrinking product never reaches.

How do I tell if a word problem needs permutations of distinct objects, or something else?

Check three things: are all the objects genuinely different from each other, is each object used at most once in an arrangement, and does the order or role matter. If yes to all three, it is this case. If objects repeat within the arrangement (like digits reused in a code), switch to the repetition-allowed formula n to the r. If some of the objects being arranged are identical to each other (like letters in a word with repeated letters), use the not-all-distinct formula instead.

Is nPn the same as n factorial, and why does nP0 equal 1?

Yes. Setting r = n in nPr = n! divided by (n-r)! gives n! divided by 0!, and since 0! is defined as 1, that is just n! — arranging all n distinct objects in every possible order. nP0 = 1 because there is exactly one way to arrange zero objects: doing nothing. Both are consistent with the shrinking-choices picture — when r = n every position gets filled down to the last object, and when r = 0 there are no positions to fill at all.

Before this

Where this leads