Q.Find the sum of all 4-digit numbers that can be formed using the digits 1,3,5,7,9.
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🔒 Start your 14-day free trial to unlock the full solution →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 …
By symmetry, each of the five digits appears the same number of times in each of the four place-value positions across all the numbers formed, which lets the total sum be built posit …
By symmetry, each of the 5 digits appears equally often in each of the 4 place-value positions; use this to find the total contribution of each position.
Digits available: 1,3,5,7,9 (sum =25). We form 4-digit numbers using 4 of these 5 digits, without repetition.
Total such numbers =5P4=5×4×3×2=120
For any fixed position (say units), fix one digit there; the remaining 3 positions are filled by arranging the other 4 digits taken 3 at a time: 4P3=24 ways. So each digit appears in each position exactly 24 times.
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Showing the 12 most recent of 27 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.If n−1P4nP4=35 then the value of n is(a) 20(b) 2(c) 10(d) 1
›Reveal solutionSolution
Write both permutation terms using factorials, cancel common factors, and solve the resulting linear equation.
nP4=(n−4)!n!,n−1P4=(n−5)!(n−1)! …
- CBSE 2026Set ANNUAL1 markQ.nPr= ______, (0≤r≤n).
›Reveal solutionSolution
nPr=(n−r)!n!.
The number of permutations of n distinct objects taken r at a time is given by choosing and arranging r positions out of n available objects without repetition: n(n−1)(n−2)⋯(n−r+1), which can be writt …
- CBSE 2025Set ANNUAL1 markMCQQ.11P0=(a) 0(b) 1(c) 11(d) 121
›Reveal solutionSolution
11P0=1.
The permutation formula is nPr=(n−r)!n!.
…
- CBSE 2025Set ANNUAL1 markMCQQ.5P0−4P0=(a) 1(b) 0(c) -1(d) 9
›Reveal solutionSolution
5P0−4P0=0.
Using nP0=1 for any n: 5P0=1 and 4P0=1.
…
- CBSE 2025Set ANNUAL1 markMCQQ.12P2=(a) 121(b) 143(c) 132(d) 12
›Reveal solutionSolution
12P2=132.
nPr=(n−r)!n!=n(n−1)(n−2)⋯(n−r+1), the product of r consecutive integers starting from n.
…
- CBSE 2025Set ANNUAL1 markMCQQ.5P3=(a) 60(b) 120(c) 180(d) 20
›Reveal solutionSolution
5P3=60.
nPr=n(n−1)(n−2)⋯(n−r+1).
5P3=5×4×3=60. …
- CBSE 2025Set ANNUAL1 markMCQQ.nP5=42×nP3, n>4 find n.(a) 12(b) 10(c) 14(d) 16
›Reveal solutionSolution
Writing both permutation terms with factorials and simplifying the ratio reduces the equation to (n−3)(n−4)=42, whose solution (with n>4) is n=10.
nP5=42×nP3
(n−5)!n!=42×(n−3)!n!
Cancel n! from both sides:
(n−5)!1=(n−3)!42
(n−5)!(n−3)!=42
…
- CBSE 2025Set ANNUAL1 markMCQQ.The number of 3 digit numbers formed by using digits 0, 1, 2, 3, 4, 5, when no digit is repeated, is:(a) 120(b) 100(c) 60(d) 720
›Reveal solutionSolution
Total arrangements of 3 digits out of 6, minus the arrangements that start with 0 (not valid 3-digit numbers).
Digits available: 0,1,2,3,4,5 (6 digits), no repetition, forming 3-digit numbers.
Total ways to arrange 3 digits chosen from 6 (ignoring the leading-digit restriction): 6P3=6×5×4=120.
…
- CBSE 2025Set sz1 markMCQQ.The value of (n−r)!n!, when n=5,r=2 is :(a) 20(b) 10(c) 30(d) 15
›Reveal solutionSolution
(n−r)!n!=nPr counts ordered arrangements; for n=5,r=2 this is 5×4=20.
By definition, nPr=(n−r)!n!.
Substitute n=5, r=2: …
- CBSE 2024Set ANNUAL1 markMCQQ.Find the number of permutations of the letters of the word 'Independence'.(a) 3!2!4!12!(b) 3!4!2!9!(c) 1!3!5!12!(d) None of these
›Reveal solutionSolution
Count the total letters and the repetition of each distinct letter, then apply the formula for permutations of a word with repeated letters: p1!p2!⋯n!.
Write out INDEPENDENCE letter by letter: I, N, D, E, P, E, N, D, E, N, C, E — that's 12 letters total.
Count each distinct letter's frequency:
- I: 1
- N: 3 (positions 2, 7, 10)
- D: 2 (positions 3, 8)
- E: 4 (positions 4, 6, 9, 12)
- P: 1
- C: 1
Check: 1+3+2+4+1+1=12 ✓.
The number of distinct permutations of a word with repeated letters is:
p1!p2!⋯pk!n! …
- CBSE 2024Set ANNUAL1 markMCQQ.If nP5 = 42 nP3, n > 4, the value of n is:(a) 10(b) 6(c) 0(d) 1
›Reveal solutionSolution
(n−3)(n−4)=42 solves to n=10 (rejecting the invalid root n=−3).
nP5=(n−5)!n!, nP3=(n−3)!n!.
nP3nP5=(n−5)!(n−3)!=(n−3)(n−4).
Given nP5=42nP3:
(n−3)(n−4)=42
n2−7n+12=42
n2−7n−30=0 …
- CBSE 2024Set sz1 markMCQQ.The value of (n−r)!n! when n=6,r=2 is equal to:(a) 6(b) 8(c) 15(d) 30
›Reveal solutionSolution
(n−r)!n! with n=6,r=2 equals 30.
The expression (n−r)!n! is exactly the formula for nPr (permutations of n things taken r at a time).
Substitute n=6,r=2: …
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