Question
If P(n − 1, 3) : P(n, 4) = 1 : 9, find n.
✓
Answer
We have,
P(n − 1, 3) : P(n, 4) = 1 : 9
$\Rightarrow \frac{\text{p}(\text{n-1,3})}{\text{P} (\text{n}, 4)}=\frac{1}{9}$
$\Rightarrow\frac{\frac{\text{(n-1)!}}{(\text{(n-1-3)!})}}{\frac{\text{n}!}{\text{(n-4)!}}}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1) }\times\text{(n-4)!}}{\text{(n-4)!}\times\text{n}!}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1)}}{\text{n!}}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1)!}}{\text{n}\times(\text{n-1})!}=\frac{1}{9}$
$\Rightarrow\frac{1}{\text{n}}=\frac{1}{9}$
$\Rightarrow\text{n}=9$
Hence, $\text{n}=9$
P(n − 1, 3) : P(n, 4) = 1 : 9
$\Rightarrow \frac{\text{p}(\text{n-1,3})}{\text{P} (\text{n}, 4)}=\frac{1}{9}$
$\Rightarrow\frac{\frac{\text{(n-1)!}}{(\text{(n-1-3)!})}}{\frac{\text{n}!}{\text{(n-4)!}}}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1) }\times\text{(n-4)!}}{\text{(n-4)!}\times\text{n}!}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1)}}{\text{n!}}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1)!}}{\text{n}\times(\text{n-1})!}=\frac{1}{9}$
$\Rightarrow\frac{1}{\text{n}}=\frac{1}{9}$
$\Rightarrow\text{n}=9$
Hence, $\text{n}=9$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = $\{$(x, y): the difference between x and y is odd, $\text{x}\in\text{A},\text{y}\in\text{B}\}$ Write R in Roster form.
→
Differentiate the following from the first principle
→$\text{e}^{\text{x}^2+1} $
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{\text{x}+3}-\sqrt{6}}{\text{x}^2-9}$
→$\lim\limits_{\text{x}\rightarrow3}\frac{\sqrt{\text{x}+3}-\sqrt{6}}{\text{x}^2-9}$
If $u=\{1,2,3,4,5,6,7,8,9,10,12,24\}$
$A=\{x: x$ is prime and $x \leq 10\}$
$B=\{x: x$ is a factor of 24$\}$
Verify the following result
i. $A - B = A \cap B^{\prime}$
ii. $(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
iii. $(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$
→$A=\{x: x$ is prime and $x \leq 10\}$
$B=\{x: x$ is a factor of 24$\}$
Verify the following result
i. $A - B = A \cap B^{\prime}$
ii. $(A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}$
iii. $(A \cap B)^{\prime}=A^{\prime} \cup B^{\prime}$
Prove that: $\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B}}=\frac {\sin\text{A+B}}{\sin\text{A-B}}$
→How many different arrangements can be made by using all the letters in the word 'MATHEMATICS'. How many of them begin with C? How many of them begin with T?
The longest side of a triangle is three times the shortest side and third side is 2 cm shorter than the longest side if the perimeter of the triangles at least 61 cm, find the minimum length of the shortest-side.
Differentiate the following function with respect to x:
→$(1+\text{x}^2)\cos\text{x}$
If $\lim\limits_{\text{x} \rightarrow 1}\frac{\text{x}^{4}-1}{\text{x}-1}=\lim\limits_{\text{x} \rightarrow{\text{k}}}\frac{\text{x}^{3}-\text{k}^{3}}{\text{x}^{2}-\text{k}^{2}}$ then find the value of K.
→Find the sum of the following geometric series:
$\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+\ ... \text{ to 8 terms;}$
→$\sqrt{2}+\frac{1}{\sqrt{2}}+\frac{1}{2\sqrt{2}}+\ ... \text{ to 8 terms;}$