$^5P_0 - {}^4P_0 =$ (a) 1 (b) 0 (c) -1 (d) 9

> > [!TLDR] > $^5P_0 - {}^4P_0 = 0$. > > Using $^nP_0 = 1$ for any $n$: $^5P_0 = 1$ and $^4P_0 = 1$. > …

⁵P₀ - ⁴P₀ = (a) 1 (b) 0 (c) -1 (d) 9 – Class 11 Mathematics

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^5P_0 - {}^4P_0 =

(a) 1

(b) 0

(d) 9

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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 = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 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 = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (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

Important

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: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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$^5P_0 - {}^4P_0 = 0$ .

Using $^nP_0 = 1$ for any $n$ : $^5P_0 = 1$ and $^4P_0 = 1$ .

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 27 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\dfrac{^nP_4}{^{n-1}P_4} = \dfrac{5}{3}$
(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.

${}^nP_4 = \dfrac{n!}{(n-4)!}, \qquad {}^{n-1}P_4 = \dfrac{(n-1)!}{(n-5)!}$ …

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CBSE 2026Set ANNUAL1 mark
${}^{n}P_r =$
›Reveal solutionSolution

$^{n}P_r = \dfrac{n!}{(n-r)!}$ .

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: $n(n-1)(n-2)\cdots(n-r+1)$ , which can be writt …

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CBSE 2025Set ANNUAL1 markMCQ
$^{11}P_0 =$
(a) 0
(b) 1
(c) 11
(d) 121
›Reveal solutionSolution

$^{11}P_0 = 1$ .

The permutation formula is $^nP_r = \dfrac{n!}{(n-r)!}$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$^5P_0 - {}^4P_0 =$
(a) 1
(b) 0
(c) -1
(d) 9
›Reveal solutionSolution

$^5P_0 - {}^4P_0 = 0$ .

Using $^nP_0 = 1$ for any $n$ : $^5P_0 = 1$ and $^4P_0 = 1$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$^{12}P_2 =$
(a) 121
(b) 143
(c) 132
(d) 12
›Reveal solutionSolution

$^{12}P_2 = 132$ .

$^nP_r = \dfrac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$ , the product of $r$ consecutive integers starting from $n$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$^5P_3 =$
(a) 60
(b) 120
(c) 180
(d) 20
›Reveal solutionSolution

$^5P_3 = 60$ .

$^nP_r = n(n-1)(n-2)\cdots(n-r+1)$ .

$^5P_3 = 5 \times 4 \times 3 = 60$ . …

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CBSE 2025Set ANNUAL1 markMCQ
${}^nP_5 = 42 \times {}^nP_3$
(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 $n=10$ .

${}^nP_5 = 42 \times {}^nP_3$

$\frac{n!}{(n-5)!} = 42 \times \frac{n!}{(n-3)!}$

Cancel $n!$ from both sides:

$\frac{1}{(n-5)!} = \frac{42}{(n-3)!}$

$\frac{(n-3)!}{(n-5)!} = 42$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The number of 3 digit numbers formed by using digits 0, 1, 2, 3, 4, 5, when no digit is repeated, is:$
(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): $^6P_3 = 6\times5\times4 = 120$ .

…

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CBSE 2025Set sz1 markMCQ
$\dfrac{n!}{(n-r)!}$
(a) 20
(b) 10
(c) 30
(d) 15
›Reveal solutionSolution

$\dfrac{n!}{(n-r)!} = {}^{n}P_r$ counts ordered arrangements; for $n=5, r=2$ this is $5\times4=20$ .

By definition, ${}^nP_r = \dfrac{n!}{(n-r)!}$ .

Substitute $n=5$ , $r=2$ : …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Find the number of permutations of the letters of the word 'Independence'.$
(a) $\dfrac{12!}{3!\,2!\,4!}$
(b) $\dfrac{9!}{3!\,4!\,2!}$
(c) $\dfrac{12!}{1!\,3!\,5!}$
(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: $\dfrac{n!}{p_1!\,p_2!\,\cdots}$ .

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:

$\frac{n!}{p_1!\,p_2!\,\cdots\,p_k!}$ …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. If nP5 = 42 nP3, n > 4, the value of n is:$
(a) 10
(b) 6
(c) 0
(d) 1
›Reveal solutionSolution

$(n-3)(n-4)=42$ solves to $n=10$ (rejecting the invalid root $n=-3$ ).

$^nP_5 = \dfrac{n!}{(n-5)!}$ , $^nP_3 = \dfrac{n!}{(n-3)!}$ .

$\dfrac{^nP_5}{^nP_3} = \dfrac{(n-3)!}{(n-5)!} = (n-3)(n-4)$ .

Given $^nP_5 = 42\,^nP_3$ :

$(n-3)(n-4) = 42$

$n^2 - 7n + 12 = 42$

$n^2 - 7n - 30 = 0$ …

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CBSE 2024Set sz1 markMCQ
$\dfrac{n!}{(n-r)!}$
(a) 6
(b) 8
(c) 15
(d) 30
›Reveal solutionSolution

$\dfrac{n!}{(n-r)!}$ with $n=6, r=2$ equals $30$ .

The expression $\dfrac{n!}{(n-r)!}$ is exactly the formula for $^nP_r$ (permutations of $n$ things taken $r$ at a time).

Substitute $n=6, r=2$ : …

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