The problem provides a ratio of two permutation expressions involving an unknown variable 'n'. We need to find the value of 'n' that satisfies this ratio.

The given ratio is: $ (2n+1)P(n–1): (2n–1)Pn = 3 : 5 $

This can be written as a fraction:

$ \frac{(2n+1)P(n–1)}{(2n–1)Pn} = \frac{3}{5} $

Recall the permutation formula: $ nPr = \frac{n!}{(n-r)!} $

Let's apply the formula to each term in the ratio:

• For $ (2n+1)P(n–1) $: Here, $ N = 2n+1 $ and $ R = n–1 $. The difference $ N - R = (2n+1) - (n–1) = 2n+1-n+1 = n+2 $. So, $ (2n+1)P(n–1) = \frac{(2n+1)!}{(n+2)!} $.
• For $ (2n–1)Pn $: Here, $ N = 2n–1 $ and $ R = n $. The difference $ N - R = (2n–1) - n = n–1 $. So, $ (2n–1)Pn = \frac{(2n–1)!}{(n–1)!} $.

Substitute these into the ratio equation:

$ \frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n–1)!}{(n–1)!}} = \frac{3}{5} $

Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator:

$ \frac{(2n+1)!}{(n+2)!} \times \frac{(n–1)!}{(2n–1)!} = \frac{3}{5} $

Now, expand the factorials to find common terms to cancel. Remember $ k! = k \times (k-1)! $

• $ (2n+1)! = (2n+1)(2n)(2n–1)! $
• $ (n+2)! = (n+2)(n+1)n(n–1)! $

Substitute these expansions into the equation:

$ \frac{(2n+1)(2n)(2n–1)!}{(n+2)(n+1)n(n–1)!} \times \frac{(n–1)!}{(2n–1)!} = \frac{3}{5} $

Cancel out the common factorial terms $ (2n–1)! $ and $ (n–1)! $ from the numerator and denominator:

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

Assuming $ n \neq 0 $ (which is required for $ Pn $ to be defined with $n \ge 1$ for the second term), we can cancel 'n' from the numerator $2n$ and the denominator $n$:

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

Now, cross-multiply to solve for n:

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

$ 10(2n+1) = 3(n^2 + n + 2n + 2) $

$ 20n + 10 = 3(n^2 + 3n + 2) $

$ 20n + 10 = 3n^2 + 9n + 6 $

Rearrange the terms to form a quadratic equation:

$ 3n^2 + 9n - 20n + 6 - 10 = 0 $

$ 3n^2 - 11n - 4 = 0 $

Solve this quadratic equation. We can factor it:

Find two numbers that multiply to $ 3 \times (-4) = -12 $ and add up to $ -11 $. These numbers are $ -12 $ and $ 1 $.

Rewrite the middle term $ -11n $ as $ -12n + n $:

$ 3n^2 - 12n + n - 4 = 0 $

Group terms and factor:

$ 3n(n - 4) + 1(n - 4) = 0 $

Factor out the common term $ (n - 4) $:

$ (n - 4)(3n + 1) = 0 $

This gives two possible solutions for n:

• $ n - 4 = 0 \implies n = 4 $
• $ 3n + 1 = 0 \implies n = -\frac{1}{3} $

Consider the domain for permutations $ NPr $. We must have $ N $ as a non-negative integer and $ R $ as a non-negative integer, with $ N \ge R $. Let's check the requirements for our specific terms:

• For $ (2n+1)P(n–1) $, we need $ 2n+1 \ge 0 $, $ n–1 \ge 0 $, and $ 2n+1 \ge n–1 $.
  • $ 2n+1 \ge 0 \implies n \ge -\frac{1}{2} $
  • $ n–1 \ge 0 \implies n \ge 1 $
  • $ 2n+1 \ge n–1 \implies n \ge -2 $
  Combining these, we need $ n \ge 1 $.
• For $ (2n–1)Pn $, we need $ 2n–1 \ge 0 $, $ n \ge 0 $, and $ 2n–1 \ge n $.
  • $ 2n–1 \ge 0 \implies n \ge \frac{1}{2} $
  • $ n \ge 0 $
  • $ 2n–1 \ge n \implies n \ge 1 $
  Combining these, we need $ n \ge 1 $.

Both permutation terms require $ n \ge 1 $. Also, for $ P(n, r) $, $n$ must typically be an integer. Of the two solutions found ($ n = 4 $ and $ n = -\frac{1}{3} $), only $ n = 4 $ is an integer and satisfies the condition $ n \ge 1 $.

Therefore, the only valid value for n is 4.

Let's verify for $ n=4 $:

$ (2(4)+1)P(4–1) = 9P3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} = 9 \times 8 \times 7 = 504 $

$ (2(4)–1)P4 = 7P4 = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840 $

The ratio is $ 504 : 840 $.

Divide both by 168: $ 504 \div 168 = 3 $ and $ 840 \div 168 = 5 $.

The ratio is $ 3:5 $, which matches the given ratio. Thus, $ n=4 $ is the correct value.

The value of n is 4.