Q.Evaluate
Concept understanding — Permutations Without Repetition
Permutations Without Repetition – The Idea of Arranging Things
Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?
You could try listing them out:
- Physics, Chemistry, Maths
- Physics, Maths, Chemistry
- Chemistry, Physics, Maths
- Chemistry, Maths, Physics
- Maths, Physics, Chemistry
- Maths, Chemistry, Physics
That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.
Why "Without Repetition"?
The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.
This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.
The Counting Logic – Why Multiply?
Let's build the arrangement step by step for 3 distinct books:
- First position: You have 3 choices (any of the 3 books).
- Second position: After placing the first book, only 2 books remain — so 2 choices.
- Third position: Only 1 book is left — so 1 choice.
Total arrangements = 3×2×1=6.
This product 3×2×1 is called 3 factorial, written as 3!.
P(n)=n!=n×(n−1)×(n−2)×⋯×2×1
For n distinct objects, the number of permutations (arrangements in order) is n!.
What If You Only Arrange Some of Them?
Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?
- First position: 5 choices
- Second position: 4 choices
- Third position: 3 choices
Total = 5×4×3=60.
This is a permutation of 5 objects taken 3 at a time, written as P(5,3) or 5P3.
P(n,r)=(n−r)!n!=n×(n−1)×⋯×(n−r+1)
Here n is the total number of distinct objects, and r is how many you are arranging. The formula works because:
- Numerator n! counts all arrangements of all n objects.
- Denominator (n−r)! removes the arrangements of the n−r objects you are not using.
Key Points to Remember
- Order matters — swapping two objects gives a different permutation.
- No repetition — each object is used at most once.
- For arranging all n objects: n!
- For arranging r out of n objects: (n−r)!n!
Common Mistake to Avoid
Do not use the permutation formula when order doesn't matter. For example, choosing 3 friends from a group of 5 to form a committee — here the order of selection is irrelevant. That's a combination, not a permutation. Permutations are for ordered arrangements (like rankings, seating orders, passwords where position matters).
Quick Examples
| Scenario | Calculation | Answer |
|---|---|---|
| Arranging 4 different trophies on a shelf | 4! | 24 |
| Number of 3-digit codes from digits 1–9 (no digit repeated) | P(9,3)=9×8×7 | 504 |
| Seating 5 people in 5 chairs | 5! | 120 |
| Assigning gold, silver, bronze medals to 8 runners | P(8,3)=8×7×6 | 336 |
The Bottom Line
Permutations without repetition answer the question: "In how many different ordered ways can I arrange a set of distinct items, using each item at most once?" The answer is always a product of decreasing integers, starting from n and going down r steps. When r=n, it's simply n!.
Permutations Without Repetition is introduced in the NCERT Class 11 Mathematics Permutations and Combinations chapter, and it is exactly the kind of topic students look up when searching "permutations formula class 11 maths" or "arrangement of distinct objects important questions". It also forms the basis for many JEE Main and state CET counting problems that ask you to arrange distinct items without repeating any of them.
Concept: Permutations Without Repetition — the factorial n! counts the number of ways to arrange n distinct objects in a sequence.
Step 1: Recall the definition: n!=n×(n−1)×(n−2)×⋯×2×1.
Step 2: Compute each factorial directly:
- 5!=5×4×3×2×1=120
- 7!=7×6×5×4×3×2×1=5040
Step 3: For part (iii), subtract: 7!−5!=5040−120=4920.
The values are 5!=120, 7!=5040, and 7!−5!=4920.
Factorials count the number of ways to arrange distinct objects in a line. 5!=120, 7!=5040, and 7!−5!=4920.
The Core Idea: What a Factorial Really Means
A factorial, written as n!, is shorthand for the product of all positive integers from 1 up to n:
n!=n×(n−1)×(n−2)×⋯×2×1
But the real power of factorials comes from permutations without repetition. Imagine you have n distinct books and you want to arrange them on a shelf. For the first position, you have n choices. After placing one, you have n−1 choices for the second spot, then n−2 for the third, and so on. By the multiplication principle, the total number of arrangements is n×(n−1)×⋯×1=n!.
This is why n! grows so fast — each new object multiplies the total by a larger number.
A common mistake is to think n! means "multiply n by itself" or to confuse it with n2. Factorials are a chain of decreasing multipliers, not repeated multiplication of the same number.
Step-by-Step Evaluation
1. Evaluate 5!
Start from 5 and multiply by every integer down to 1:
5!=5×4×3×2×1
Work through it in stages:
- 5×4=20
- 20×3=60
- 60×2=120
- 120×1=120
So 5!=120.
You can also think of 5! as 5×4!. Since 4!=24, we get 5×24=120. This recursive property — n!=n×(n−1)! — is extremely useful in combinatorics.
2. Evaluate 7!
Similarly:
7!=7×6×5×4×3×2×1
Multiply step by step:
- 7×6=42
- 42×5=210
- 210×4=840
- 840×3=2520
- 2520×2=5040
- 5040×1=5040
Thus 7!=5040.
Notice how quickly the numbers grow: 5! is only 120, but 7! is over 5000. Adding just two more factors (6 and 7) multiplied the result by 42.
3. Evaluate 7!−5!
Now we simply subtract the two results:
7!−5!=5040−120=4920
You cannot "factor" the subtraction in a simple way like (7−5)! — that would be 2!=2, which is completely wrong. The factorial operation does not distribute over subtraction.
The values are 5!=120, 7!=5040, and 7!−5!=4920.
Showing the 12 most recent of 18 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.Let S be the set of all 3-digit numbers containing the digits 3, 5 and 7 without repetition. The sum of all numbers in S is (A) 3330 (B) 2220 (C) 4590 (D) 1110 (E) 4440
›Reveal solutionSolution
By symmetry each place-value column sums to 2×(3+5+7)=30, giving 30×111=3330.
There are 3!=6 numbers using digits 3,5,7 without repetition. By symmetry each digit appears in each of the three positions in exactly 2 of the numbers. The sum of the digits is 3+5+7=15, so the total contribution of each position is 2×15=30 units of that place value.
Hence the grand total is
30×(100+10+1)=30×111=3330.
✓Final answerThe correct option is (A).
- KEAM 2026Set eng-2026-04174 marksMCQQ.There are 3 boys and 4 girls in a group. The number of ways they can sit in a row so that between any two boys there is a girl and between any two girls there is a boy, is (A) 88 (B) 96 (C) 124 (D) 144 (E) 288
›Reveal solutionSolution
The seating must strictly alternate; with 4 girls and 3 boys the pattern is fixed as GBGBGBG, giving 4!3!=144.
The conditions "between any two boys there is a girl" and "between any two girls there is a boy" together force a strictly alternating seating. With 3 boys and 4 girls in 7 seats, the only alternating arrangement is
GBGBGBG,
i.e. girls in the four odd positions and boys in the three even positions.
The girls can be arranged in 4! ways and the boys in 3! ways:
4!×3!=24×6=144.
✓Final answerThe correct option is (D).
- KEAM 2026Set eng-2026-04174 marksMCQQ.The number of arrangements of the letters of the word BANANA so that the arrangement starts and ends with the same letter, is (A) 12 (B) 16 (C) 18 (D) 20 (E) 24
›Reveal solutionSolution
Only A or N can occupy both ends; counting each case gives 12+4=16.
BANANA has letters A(×3),N(×2),B(×1). For an arrangement to start and end with the same letter, that letter must appear at least twice, so it is A or N.
- Both ends A: two A's used at the ends; the middle four positions hold {A,N,N,B}, arrangeable in 2!4!=12 ways.
- Both ends N: both N's used; the middle holds {A,A,A,B}, arrangeable in 3!4!=4 ways.
Total =12+4=16.
✓Final answerThe correct option is (B).
- KEAM 2026Set eng-2026-04204 marksMCQQ.If nP5=6720 and (n−1)P4=840, then nP3 is equal to (A) 346 (B) 348 (C) 396 (D) 376 (E) 336
›Reveal solutionSolution
nP5=(n−5)!n! and n−1P4=(n−5)!(n−1)!, so their ratio is n. Thus n=6720/840=8, giving nP3=8⋅7⋅6=336.
Use the permutation formulas:
nP5=(n−5)!n!,n−1P4=(n−5)!(n−1)!.
Dividing:
n−1P4nP5=(n−1)!n!=n=8406720=8.
So n=8, and
nP3=8⋅7⋅6=336.
✓Final answerThe correct option is (E).
- KEAM 2025Set eng-2025-04234 marksMCQQ.If 11Pr=7920, then the value of r is equal to (A) 7 (B) 6 (C) 5 (D) 4 (E) 3
›Reveal solutionSolution
11109*8 = 7920, so r = 4.
Concept and Intuition
The permutation P(n,r) is a product of r descending factors starting at n. Matching 7920 to such a descending product identifies r.
Step-by-Step Solution
- P(11, r) = 1110...*(11-r+1).
- 1110 = 110; 1109 = 990; 990*8 = 7920.
- That used four factors, so r = 4.
Common Mistakes
- Counting the number of factors incorrectly and reporting r = 3 or 5.
✓Final answerThe correct option is (D) — 4.
ANSWER: D
- KEAM 2025Set eng-2025-04234 marksMCQQ.Let A={0,2,4,6,8}. The number of 5-digit numbers that can be formed using the digits in A without replacement, is (A) 120 (B) 96 (C) 88 (D) 64 (E) 32
›Reveal solutionSolution
From 5! total arrangements we remove the 4! that start with 0, leaving 96.
Concept and Intuition
Using all five distinct digits gives 5! orderings, but a valid 5-digit number cannot begin with 0. Subtract the arrangements with 0 in the leading position.
Step-by-Step Solution
- Total arrangements of {0,2,4,6,8} = 5! = 120.
- Arrangements with 0 leading = 4! = 24.
- Valid numbers = 120 - 24 = 96.
Common Mistakes
- Reporting 120 and forgetting to exclude leading zeros.
✓Final answerThe correct option is (B) — 96.
ANSWER: B
- KEAM 2025Set eng-2025-04264 marksMCQQ.25 distinct objects are divided into 5 groups and each group consists of exactly 5 objects. Then the number of ways of forming such groups, is (A) (5!)525! (B) 5!25! (C) (5!)625! (D) (5!)425! (E) (5!)325!
›Reveal solutionSolution
Groups are of equal size and unlabelled, so divide by (5!)5 for internal order and by 5! for the interchange of the identical groups: (5!)625!.
Reasoning. Arrange all 25 objects: 25!. Each group of 5 is unordered ⇒ divide by (5!)5. The five groups themselves are of the same size and unlabelled ⇒ divide by 5!.
(5!)5⋅5!25!=(5!)625!.
✓Final answerThe correct option is (C).
- KEAM 2025Set eng-2025-04274 marksMCQQ.The number of integers greater than 7000 using 2,4,6,7,8 without repetition, is (A) 168 (B) 336 (C) 196 (D) 256 (E) 512
›Reveal solutionSolution
Count 4-digit numbers starting with 7 or 8 (48) plus all 5-digit arrangements (120): total 168.
The available digits are 2,4,6,7,8 (all distinct, no repetition). A number using these digits and greater than 7000 is either a 4-digit or a 5-digit number.
4-digit numbers >7000: the thousands digit must be ≥7, so it is 7 or 8 (2 choices). The remaining three places are filled from the other 4 digits:
2×(4×3×2)=2×24=48.
5-digit numbers: any arrangement of all five digits is at least 20000>7000, so all count:
5!=120.
Total =48+120=168.
✓Final answerThe correct option is (A).
- KEAM 2025Set eng-2025-04274 marksMCQQ.Five digit number is formed using the digits 0,1,2,3,4 and 5 without repetitions. Number of five digit numbers which are divisible by 10 is (A) 360 (B) 240 (C) 120 (D) 480 (E) 520
›Reveal solutionSolution
Fix the last digit as 0; the other four positions use the digits 1–5: P(5,4)=120.
A 5-digit number formed from {0,1,2,3,4,5} (no repetition) is divisible by 10 iff its units digit is 0.
Fixing the units digit as 0, the remaining four positions are filled by choosing and arranging 4 of the remaining 5 digits {1,2,3,4,5}:
5×4×3×2=120.
Since 0 occupies the units place, the leading digit is automatically non-zero, so all 120 are valid 5-digit numbers.
✓Final answerThe correct option is (C).
- KEAM 2025Set eng-2025-04284 marksMCQQ.If n(A)=8, then the number of subsets of A which contain 2 or 6 elements is (A) 24 (B) 28 (C) 48 (D) 56 (E) 216
›Reveal solutionSolution
Subsets with exactly 2 elements: (28)=28; with exactly 6: (68)=28. Total =56.
With n(A)=8, the number of subsets containing exactly 2 elements is (28)=28, and exactly 6 elements is (68)=(28)=28.
These two families are disjoint, so the total is
28+28=56.
✓Final answerThe correct option is (D).
- KEAM 2025Set eng-2025-04284 marksMCQQ.Four digit numbers are formed using 0, 3, 4, 5, 9, 8 without repetitions. Then the number of such 4 digits numbers is (A) 270 (B) 300 (C) 320 (D) 400 (E) 450
›Reveal solutionSolution
6 available digits {0,3,4,5,9,8}, no repetition. Leading digit can't be 0 (5 ways); fill remaining 3 places from the other 5 digits (5⋅4⋅3). Total =300.
For a 4-digit number the first (thousands) place cannot be 0, so it has 5 choices.
The remaining three places are filled from the remaining 5 digits without repetition:
5×4×3=60.
Total =5×60=300.
✓Final answerThe correct option is (B).
- KEAM 2025Set eng-2025-04284 marksMCQQ.A bag contains 5 red balls, 4 black balls, and 3 white balls. Then the number of ways of selecting three balls at random that contains at least one white ball is (A) 220 (B) 210 (C) 180 (D) 136 (E) 74
›Reveal solutionSolution
At least one white = (all selections) − (no white) =(312)−(39)=220−84=136.
There are 5+4+3=12 balls. Choosing 3 in all: (312)=220.
Selections with no white ball come from the 9 non-white balls: (39)=84.
At least one white =220−84=136.
✓Final answerThe correct option is (D).
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