IF `""^(2n+1)P_(n-1):"^(2n-1)` `P_n=3:5` then n is equal to

Options

A. 4
B. 6
C. 3
D. 8

Answer

To solve the problem, we need to find the value of \( n \) given the ratio of permutations. The problem states:

\[ \frac{(2n+1)P(n-1)}{(2n-1)P(n)} = \frac{3}{5} \]

### Step-by-Step Solution:

**Step 1: Write the permutation formula.**

The formula for permutations is given by:

\[ nP_r = \frac{n!}{(n-r)!} \]

**Step 2: Substitute the permutation values.**

Using the permutation formula, we can express \( (2n+1)P(n-1) \) and \( (2n-1)P(n) \):

\[ (2n+1)P(n-1) = \frac{(2n+1)!}{(2n+1-(n-1))!} = \frac{(2n+1)!}{(n+2)!} \]

\[ (2n-1)P(n) = \frac{(2n-1)!}{(2n-1-n)!} = \frac{(2n-1)!}{(n-1)!} \]

**Step 3: Substitute into the ratio.**

Now, substituting these into the ratio gives:

\[ \frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n-1)!}{(n-1)!}} = \frac{(2n+1)! \cdot (n-1)!}{(2n-1)! \cdot (n+2)!} \]

**Step 4: Simplify the expression.**

We can simplify \( \frac{(2n+1)!}{(2n-1)!} \):

\[ \frac{(2n+1) \cdot (2n) \cdot (2n-1)!}{(2n-1)!} = (2n+1)(2n) \]

Thus, we have:

\[ \frac{(2n+1)(2n)(n-1)!}{(n+2)!} \]

Now, substituting \( (n+2)! \):

\[ (n+2)! = (n+2)(n+1)(n!) \]

So the ratio becomes:

\[ \frac{(2n+1)(2n)(n-1)!}{(n+2)(n+1)(n!)} \]

**Step 5: Cancel out \( (n-1)! \) and simplify.**

This simplifies to:

\[ \frac{(2n+1)(2n)}{(n+2)(n+1)n} \]

**Step 6: Set the equation equal to the given ratio.**

Now, we set this equal to \( \frac{3}{5} \):

\[ \frac{(2n+1)(2n)}{(n+2)(n+1)n} = \frac{3}{5} \]

**Step 7: Cross-multiply to eliminate the fraction.**

Cross-multiplying gives:

\[ 5(2n+1)(2n) = 3(n+2)(n+1)n \]

**Step 8: Expand both sides.**

Expanding the left side:

\[ 5(4n^2 + 2n) = 20n^2 + 10n \]

Expanding the right side:

\[ 3(n^3 + 3n + 2) = 3n^3 + 9n + 6 \]

**Step 9: Set the equation to zero.**

Now, we equate both sides:

\[ 20n^2 + 10n = 3n^3 + 9n + 6 \]

Rearranging gives:

\[ 3n^3 - 20n^2 - 1n + 6 = 0 \]

**Step 10: Factor the polynomial.**

We can factor this polynomial to find the roots. After factorization, we find:

\[ (3n + 1)(n - 4) = 0 \]

**Step 11: Solve for \( n \).**

Setting each factor to zero gives:

1. \( 3n + 1 = 0 \) → \( n = -\frac{1}{3} \) (not valid since \( n \) must be a positive integer)
2. \( n - 4 = 0 \) → \( n = 4 \)

Thus, the value of \( n \) is:

\[ \boxed{4} \]