Update solution.md (#36)

change 2!/3^n to 1!/3^n as you'd get 14/3 as the answer and that makes no sense.

Author: BradleyNgu

src/content/questions/comp2804/2019-fall-midterm/10/solution.md

@@ -8,7 +8,7 @@
 
 <ul>
     <li>$f(n) = 7 \cdot \frac{n!}{3^n}$ <br/>
-    $f(1) = 7 \cdot \frac{2!}{3^1} = 7 \cdot \frac{1}{3} = \frac{7}{3}$
+    $f(1) = 7 \cdot \frac{1!}{3^1} = 7 \cdot \frac{1}{3} = \frac{7}{3}$
     <li>$f(n) = 7^n \cdot \frac{n!}{3^n}$ <br/>
     $f(1) = 7^1 \cdot \frac{2!}{3^1} = 7 \cdot \frac{2}{3} = \frac{14}{3}$
     <li>$f(n) = 7 \cdot \frac{(n + 1)!}{3^n}$ <br/>