To solve the given equation [tex]\(\frac{n P_3}{n C_3} = 6\)[/tex], we need to use the definitions of permutations [tex]\(P\)[/tex] and combinations [tex]\(C\)[/tex].
First, recall the formula for permutations and combinations:
- Permutations of selecting 3 elements from [tex]\(n\)[/tex] elements is denoted as [tex]\(n P_3\)[/tex] and is given by:
[tex]\[
n P_3 = \frac{n!}{(n-3)!}
\][/tex]
- Combinations of selecting 3 elements from [tex]\(n\)[/tex] elements is denoted as [tex]\(n C_3\)[/tex] and is given by:
[tex]\[
n C_3 = \frac{n!}{3!(n-3)!}
\][/tex]
Next, substitute these formulas into the given equation:
[tex]\[
\frac{n P_3}{n C_3} = \frac{\frac{n!}{(n-3)!}}{\frac{n!}{3!(n-3)!}} = 6
\][/tex]
Simplify the fraction:
[tex]\[
\frac{\frac{n!}{(n-3)!}}{\frac{n!}{3!(n-3)!}} = \frac{n!}{(n-3)!} \times \frac{3!(n-3)!}{n!} = \frac{3!}{1} = 3!
\][/tex]
Since [tex]\(3!\)[/tex] (3 factorial) is equal to 6:
[tex]\[
3! = 6
\][/tex]
As this holds true for [tex]\(3! = 6\)[/tex], we need to equate both sides:
[tex]\[
6 = 6
\][/tex]
However, this equation holds true universally and does not yield a specific value for [tex]\(n\)[/tex]. Therefore, there is no particular value of [tex]\(n\)[/tex] that satisfies our original equation:
[tex]\[
\frac{n P_3}{n C_3} = 6
\][/tex]
This means there is no solution for [tex]\(n\)[/tex] that can satisfy the given equation. Thus, the result is:
[tex]\[
\boxed{\text{No solution}}
\][/tex]
