Question
If ${ }^{2 n+1} P_{n-1}:{ }^{2 n-1} P_{n}=3: 5$, then find the value of $n$.
Step 1
We know that $nPr = \frac{n!}{(n-r)!}$. So, we can write the given equation as $\frac{(2n+1)!}{(2n+1-(n-1))!} : \frac{(2n-1)!}{(2n-1-n)!} = 3:5$. Show more…
Show all steps
Your feedback will help us improve your experience
Aman Gupta and 67 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Solve for the value of $n$. $$\frac{n !}{(n+2) !} \cdot \frac{(n+3) !}{n(n-1) !}=5$$
Principles of Counting and Theories of Probability
Exercise 5
For what value of $n$ is $_{n+1} P_{3}={ }_{n} P_{4} ?$
$$\text { Find } 3 n^{2} \text { if } n(n+5)=-4$$
Factoring
Solving Equations
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD