Find the value of $n$ such that (i) $^nP_4 = 20\ ^nP_2$, $n > 3$ (ii) $\dfrac{^nP_4}{^{n-1}P_4} = \dfrac{5}{3}$, $n > 4$

> [!TLDR] Expand each permutation in factorial form, cancel common factors, and solve the resulting polynomial/linear equation for $n$, keeping only the root satisfying the given restriction. > [!FORMULA] > $^{n}P_r = \dfrac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$ ($r$ factors). **(i)** $^{n}P_4 = 20\,{}^{n}P_2$, with $n>3$. 1. Write out both sides: $^{n}P_4 = n(n-1)(n-2)(n-3)$ and $^{n}P_2 = n(n-1)$. 2. Equation: $n(n-1)(n-2)(n-3) = 20\,n(n-1)$. 3. Since $n>3$, $n(n-1)\neq 0$; divide both sides by $n(n-1)$: $(n-2)(n-3) = 20$. 4. Expand: $n^2 - 5n + 6 = 20 \Rightarrow n^2 - 5n - 14 = 0$. 5. Factor: $(n-7)(n+2) = 0 \Rightarrow n = 7$ or $n = -2$. 6. Since $n>3$, reject $n=-2$; so $n = 7$. 7. Check: $^{7}P_4 = 7\times6\times5\times4 = 840$; $^{7}P_2 = 7\times6 = 42$; $20\times42 = 840$ ✓. **(ii)** $\dfrac{{}^{n}P_4}{{}^{n-1}P_4} = \dfrac{5}{3}$, with $n>4$. 8. $^{n}P_4 = n(n-1)(n-2)(n-3)$ and $^{n-1}P_4 = (n-1)(n-2)(n-3)(n-4)$. 9. Ratio: $\dfrac{n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)(n-4)} = \dfrac{n}{n-4}$ (cancel the common factor $(n-1)(n-2)(n-3)$, valid since $n>4$). 10. So $\dfrac{n}{n-4} = \dfrac{5}{3} \Rightarrow 3n = 5(n-4) = 5n - 20 \Rightarrow -2n = -20 \Rightarrow n = 10$. 11. Check: $n=10>4$ ✓. $^{10}P_4 = 10\times9\times8\times7=5040$; $^{9}P_4=9\times8\times7\times6=3024$; $5040/3024 = 5/3$ ✓. > [!ANSWER] (i) $n=7$ (ii) $n=10$

Find the value of n such that (i) ⁿP₄ = 20ⁿP₂, n > – Class 11 Applied Mathematics

Worked Examples · Example 17

n

(i)

^nP_4 = 20\ ^nP_2

n > 3

(ii)

\dfrac{^nP_4}{^{n-1}P_4} = \dfrac{5}{3}

n > 4

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Concept understanding — Permutations

Permutations: Arranging Things in Order

Imagine you have three different books — a Physics book, a Chemistry book, and a Maths book — and you want to place them on a shelf. How many different ways can you arrange them?

You could put Physics first, then Chemistry, then Maths. Or Physics, Maths, Chemistry. Or Chemistry, Physics, Maths. And so on. Each distinct ordering is called a permutation of the three books.

Note

A permutation is simply an arrangement of objects in a specific order. The key word is order — changing the order gives a different permutation.

Building the Intuition: The Multiplication Principle

Let's count the arrangements for those three books. For the first position on the shelf, you have 3 choices (any of the three books). Once you place one, for the second position you have 2 choices left. For the third position, only 1 choice remains.

So the total number of arrangements is:

3 \times 2 \times 1 = 6

This is the multiplication principle at work: the number of ways to do a sequence of tasks is the product of the number of choices at each step.

The General Formula

Now suppose you have $n$ distinct objects and you want to arrange all of them in a row. By the same logic:

1st position: $n$ choices
2nd position: $n-1$ choices
3rd position: $n-2$ choices
…
Last position: $1$ choice

The total number of permutations of $n$ distinct objects taken all at once is:

n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

This product has a special name — factorial — written as $n!$ (read "n factorial").

n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For example, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ .

Important

By convention, $0! = 1$ . This is not a random rule — it makes formulas work consistently (there is exactly one way to arrange zero objects: do nothing).

What If You Don't Arrange All Objects?

Often you have $n$ objects but only want to arrange $r$ of them (where $r \leq n$ ). For instance, from 10 students, how many ways can you choose a first, second, and third prize winner?

1st prize: 10 choices
2nd prize: 9 choices
3rd prize: 8 choices

Total: $10 \times 9 \times 8 = 720$

In general, the number of permutations of $n$ distinct objects taken $r$ at a time is:

P(n, r) = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)

This product has exactly $r$ factors, starting from $n$ and decreasing.

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

Check: $P(10, 3) = \frac{10!}{7!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720$ . The formula works.

A Quick Example

Problem: How many 4-letter words can be formed from the letters of the word "MATHS" without repetition?

Solution: You have 5 distinct letters, and you want to arrange any 4 of them in order. So:

P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 \times 1 = 120

Tip

When $r = n$ , the formula $P(n, n) = \frac{n!}{0!} = n!$ matches our earlier result — arranging all objects.

The Core Idea

Permutations are about ordered arrangements. Whenever the sequence matters — seating arrangements, prize distributions, passwords, number plates — you are counting permutations. The factorial and the $P(n, r)$ formula are just compact ways to write the multiplication principle.

Final answer: The number of permutations of $n$ distinct objects taken $r$ at a time is $\mathbf{P(n, r) = \frac{n!}{(n-r)!}}$ , and when $r = n$ , it simplifies to $\mathbf{n!}$ .

Expand each permutation in factorial form, cancel common factors, and solve the resulting polynomial/linear equation for $n$ , keeping only the root satisfying the given restriction.

[!FORMULA]

$^{n}P_r = \dfrac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$ ( $r$ factors).

(i) $^{n}P_4 = 20\,{}^{n}P_2$ , with $n>3$ .

Write out both sides: $^{n}P_4 = n(n-1)(n-2)(n-3)$ and $^{n}P_2 = n(n-1)$ .
Equation: $n(n-1)(n-2)(n-3) = 20\,n(n-1)$ .
Since $n>3$ , $n(n-1)\neq 0$ ; divide both sides by $n(n-1)$ : $(n-2)(n-3) = 20$ .
Expand: $n^2 - 5n + 6 = 20 \Rightarrow n^2 - 5n - 14 = 0$ .
Factor: $(n-7)(n+2) = 0 \Rightarrow n = 7$ or $n = -2$ .
Since $n>3$ , reject $n=-2$ ; so $n = 7$ .
Check: $^{7}P_4 = 7\times6\times5\times4 = 840$ ; $^{7}P_2 = 7\times6 = 42$ ; $20\times42 = 840$ ✓.

(ii) $\dfrac{{}^{n}P_4}{{}^{n-1}P_4} = \dfrac{5}{3}$ , with $n>4$ .

8. $^{n}P_4 = n(n-1)(n-2)(n-3)$ and $^{n-1}P_4 = (n-1)(n-2)(n-3)(n-4)$ .

9. Ratio: $\dfrac{n(n-1)(n-2)(n-3)}{(n-1)(n-2)(n-3)(n-4)} = \dfrac{n}{n-4}$ (cancel the common factor $(n-1)(n-2)(n-3)$ , valid since $n>4$ ).

10. So $\dfrac{n}{n-4} = \dfrac{5}{3} \Rightarrow 3n = 5(n-4) = 5n - 20 \Rightarrow -2n = -20 \Rightarrow n = 10$ .

11. Check: $n=10>4$ ✓. $^{10}P_4 = 10\times9\times8\times7=5040$ ; $^{9}P_4=9\times8\times7\times6=3024$ ; $5040/3024 = 5/3$ ✓.

✓Final answer

n=7

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

CBSE 2020Set 2A2 marks
${}^{n}P_7 = 42\,{}^{n}P_5$
›Reveal solutionSolution

Write $^nP_7$ and $^nP_5$ using factorials, take the ratio to get a quadratic in $n$ , and solve.

$^nP_7 = \frac{n!}{(n-7)!}, \qquad ^nP_5 = \frac{n!}{(n-5)!}$

Given $^nP_7 = 42\, ^nP_5$ :

$\frac{n!}{(n-7)!} = 42\cdot\frac{n!}{(n-5)!}$

$\frac{(n-5)!}{(n-7)!} = 42$

$(n-5)(n-6) = 42$

Expand and solve:

$n^2 - 11n + 30 = 42 \implies n^2-11n-12=0 \implies (n-12)(n+1)=0$

So $n=12$ or $n=-1$ . Since $n$ must be a positive integer with $n\ge7$ (for $^nP_7$ to be defined), we reject $n=-1$ .

✓Final answer
$n = 12$
CBSE 2018Set 2A2 marks
$Q. Find the number of ways of arranging the letters of the word "MATHEMATICS".$
›Reveal solutionSolution

Divide the total permutations of all 11 letters by the repeated-letter corrections for M, A, and T (each repeated twice).

The word "MATHEMATICS" has $11$ letters: M, A, T, H, E, M, A, T, I, C, S.

Letter frequencies: M:2, A:2, T:2, H:1, E:1, I:1, C:1, S:1 (total $2+2+2+1+1+1+1+1=11$ ✓).

If all 11 letters were distinct, there would be $11!$ arrangements. Since M, A, and T each repeat twice, we divide by $2!$ for each repeated letter to remove the overcounting of identical arrangements:

$\text{Number of arrangements} = \dfrac{11!}{2!\,2!\,2!} = \dfrac{39916800}{8} = 4989600$

✓Final answer
$4989600$

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