Key Concept: Combinations Formula

The number of ways to choose $r$ distinct items from a set of $n$ distinct items, without regard to the order of selection, is given by the combination formula: ${ }^{n} C_r = \frac{n!}{r!(n-r)!}$ where $n$ is a non-negative integer, $r$ is a non-negative integer, and $0 \le r \le n$. It's important to remember that $n$ must be greater than or equal to $r$. If $n < r$, the combination is undefined (or considered 0 depending on context, but typically undefined in direct calculation like this).

Setting Up the Equation

We are given the ratio of two combination terms: ${}^{2 n} C_{3}:{ }^{n} C_{3}=10: 1$. This can be written as a fraction: $\frac{{ }^{2 n} C_3}{{ }^{n} C_3} = \frac{10}{1}$

Expanding and Simplifying Combinations

Now, we apply the combination formula to both the numerator and the denominator. For the numerator, ${}^{2 n} C_3$, we have $n_{num} = 2n$ and $r_{num} = 3$: ${ }^{2 n} C_3 = \frac{(2n)!}{3!((2n)-3)!}$ For the denominator, ${}^{n} C_3$, we have $n_{den} = n$ and $r_{den} = 3$: ${ }^{n} C_3 = \frac{n!}{3!(n-3)!}$

Substitute these into our ratio equation: $\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = \frac{10}{1}$ To simplify the complex fraction, we can multiply the numerator by the reciprocal of the denominator: $\frac{(2n)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n!} = 10$ Notice that $3!$ appears in both the numerator and the denominator, so they cancel out: $\frac{(2n)!}{(2n-3)!} \times \frac{(n-3)!}{n!} = 10$

Next, we expand the factorials. Remember that $k! = k \times (k-1) \times \dots \times 1$. We can also write $k! = k \times (k-1)!$ or $k! = k \times (k-1) \times (k-2)!$, and so on. This allows us to simplify ratios of factorials: $\frac{(2n)!}{(2n-3)!} = \frac{2n \times (2n-1) \times (2n-2) \times (2n-3)!}{(2n-3)!} = 2n(2n-1)(2n-2)$ And similarly: $\frac{n!}{(n-3)!} = \frac{n \times (n-1) \times (n-2) \times (n-3)!}{(n-3)!} = n(n-1)(n-2)$ So our equation becomes: $\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 10$

Common Mistake Alert: Before simplifying further, ensure that $n \ne 0$, $n \ne 1$, $n \ne 2$. From the condition $n \ge r$, for ${}^{n} C_3$ to be defined, $n$ must be at least $3$. Also, for ${}^{2n} C_3$ to be defined, $2n$ must be at least $3$, which means $n$ must be at least $1.5$. Combining these, $n$ must be an integer greater than or equal to 3. This means we can safely divide by $n$, $(n-1)$, and $(n-2)$.

We can factor out a 2 from $(2n-2)$ in the numerator: $\frac{2n(2n-1) \times 2(n-1)}{n(n-1)(n-2)} = 10$ Now, we can cancel out common terms $n$ and $(n-1)$ from the numerator and denominator: $\frac{4(2n-1)}{n-2} = 10$

Solving for 'n'

Now we solve the algebraic equation for $n$. Multiply both sides by $(n-2)$: $4(2n-1) = 10(n-2)$ Distribute the numbers on both sides: $8n - 4 = 10n - 20$ Rearrange the terms to solve for $n$. Subtract $8n$ from both sides: $-4 = 2n - 20$ Add $20$ to both sides: $16 = 2n$ Divide by $2$: $n = 8$

Let's double-check the validity of $n=8$. For ${}^{n} C_3 = {}^{8} C_3$, $n=8 \ge r=3$, which is valid. For ${}^{2n} C_3 = {}^{16} C_3$, $2n=16 \ge r=3$, which is valid. Thus, $n=8$ is the correct value.

(Self-Correction/Common Mistake): In the original solution, a quadratic equation was derived and solved, yielding $n=1$ and $n=8$. The original solution correctly identifies $n=1$ as "not valid" without a clear explanation. Let's explicitly explain why $n=1$ is not valid: If $n=1$, then the term ${}^{n} C_3$ becomes ${}^{1} C_3$. According to the definition of combinations, $n$ must be greater than or equal to $r$. Here, $n=1$ and $r=3$, so $n < r$. Therefore, ${}^{1} C_3$ is undefined, making $n=1$ an invalid solution for this problem. This step is crucial for understanding the domain of validity for combinatorial problems.

Calculating the Required Ratio

The problem asks for the ratio $(n^2+3n) : (n^2-3n+4)$. Substitute the valid value $n=8$ into this expression: $\frac{n^2+3n}{n^2-3n+4} = \frac{(8)^2 + 3(8)}{(8)^2 - 3(8) + 4}$ Calculate the numerator: $8^2 + 3(8) = 64 + 24 = 88$ Calculate the denominator: $8^2 - 3(8) + 4 = 64 - 24 + 4 = 40 + 4 = 44$ So, the ratio is: $\frac{88}{44} = 2$ This can be expressed as a ratio $2:1$.

Summary and Key Takeaway

This problem effectively tests your understanding of the combinations formula and the conditions for its validity. The key steps involved expanding the combination terms using factorials, simplifying the resulting algebraic expression, and critically, verifying the validity of the solutions for $n$ within the context of combinations (i.e., $n \ge r$). Always remember to check the domain of validity for $n$ in combination problems, as algebraic solutions might yield values that are not combinatorially meaningful.