Question:

Step 1: Set up the probability formula. For the $n$-th card drawn to be the second ace, the first $n-1$ cards must contain exactly one ace, and the $n$-th card must be an ace. The probability of drawing exactly one ace in the first $n-1$ cards is given by the ratio of favorable combinations to total combinations: $$P(\text{1 ace in first } n-1 \text{ cards}) = \frac{\binom{4}{1}\binom{48}{n-2}}{\binom{52}{n-1}}$$ Given that one ace has been drawn and $n-1$ cards are already out, there are 3 aces remaining and $52-(n-1)$ cards total remaining. The probability that the $n$-th card is an ace is: $$P(\text{n-th card is ace} | \text{1 ace in first } n-1 \text{ cards}) = \frac{3}{52-(n-1)}$$ Therefore, the probability $P(N=n)$ is the product of these two probabilities: $$P(N=n) = \frac{\binom{4}{1}\binom{48}{n-2}}{\binom{52}{n-1}} \cdot \frac{3}{52-(n-1)}$$ Step 2: Simplify the expression. $$P(N=n) = \frac{4 \cdot \binom{48}{n-2}}{\binom{52}{n-1}} \cdot \frac{3}{53-n}$$ $$P(N=n) = \frac{12}{53-n} \cdot \frac{\frac{48!}{(n-2)!(48-(n-2))!}}{\frac{52!}{(n-1)!(52-(n-1))!}}$$ $$P(N=n) = \frac{12}{53-n} \cdot \frac{48!}{(n-2)!(50-n)!} \cdot \frac{(n-1)!(53-n)!}{52!}$$ Expanding the factorials to simplify: $$P(N=n) = \frac{12}{53-n} \cdot \frac{48!}{(n-2)!(50-n)!} \cdot \frac{(n-1)(n-2)! (53-n)(52-n)(51-n)(50-n)!}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48!}$$ Canceling common terms: $$P(N=n) = \frac{12(n-1)(52-n)(51-n)}{52 \cdot 51 \cdot 50 \cdot 49}$$ Step 3: Express in the form $\frac{1}{k}(n-a)(n-b)(n-c)$. To match the given form $P(N=n) = \frac{1}{k}(n-a)(n-b)(n-c)$ with $a > b > c$, we rewrite the factors $(52-n)$ and $(51-n)$: $$(52-n) = -(n-52)$$ $$(51-n) = -(n-51)$$ Substituting these into the expression for $P(N=n)$: $$P(N=n) = \frac{12}{52 \cdot 51 \cdot 50 \cdot 49} (n-1) (-(n-52)) (-(n-51))$$ $$P(N=n) = \frac{12}{52 \cdot 51 \cdot 50 \cdot 49} (n-1) (n-52) (n-51)$$ Comparing this with $\frac{1}{k}(n-a)(n-b)(n-c)$, the factors are $(n-1)$, $(n-51)$, and $(n-52)$. To satisfy the condition $a > b > c$, we assign the values: $$a=52, \quad b=51, \quad c=1$$ This ordering $52 > 51 > 1$ is consistent with the condition. The constant $k$ is determined by: $$k = \frac{52 \cdot 51 \cdot 50 \cdot 49}{12}$$ Step 4: Evaluate the properties of $a, b, c, k$. 1. The value of $a$ is 52. This is consistent with our derivation ($a=52$). 2. The value of $b+c$ is 52. This is consistent with our derivation ($b+c = 51+1 = 52$). 3. The value of $a+c$ is 52. This is not consistent with our derivation ($a+c = 52+1 = 53$). 4. $17$ is a factor of $k$. $$k = \frac{52 \cdot 51 \cdot 50 \cdot 49}{12}$$ Since $51 = 3 \times 17$, $k$ contains a factor of $17$. Thus, $17$ is a factor of $k$. This is consistent with our derivation.