Question

If ${}^1P_1 + 2 \cdot {}^2P_2 + 3 \cdot {}^3P_3 + \dots + 15 \cdot {}^{15}P_{15} = {}^qP_r - s$, where $0 \leq s \leq 1$, then $q + s + {}^qC_{r - s}$ is equal to _________.

Hint:

The identity $\sum_{k=1}^n k \cdot k! = (n + 1)! - 1$ is a standard result in permutations and series.

Answer 1

Correct Answer: 136

Step 1: Understanding the Concept:
We use the general term of the summation: $k \cdot {}^k P_k = k \cdot k!$. We can express this in a telescopic form to evaluate the sum.
Step 2: Detailed Explanation:
Note that $k \cdot k! = (k + 1 - 1)k! = (k + 1)! - k!$.
Sum $= \sum_{k=1}^{15} ((k + 1)! - k!) = (2! - 1!) + (3! - 2!) + \dots + (16! - 15!)$.
Sum $= 16! - 1! = 16! - 1$.
Given this is equal to ${}^q P_r - s$.
Since $0 \leq s \leq 1$, we have $s = 1$ and ${}^q P_r = 16!$.
This gives $q = 16, r = 16$.
Now calculate $q + s + {}^q C_{r - s}$:
\[ 16 + 1 + {}^{16}C_{16 - 1} = 17 + {}^{16}C_{15} = 17 + 16 = 33 \]
Step 3: Final Answer:
The value is 33.