This is the question:
Given the letters of the word "CORONA", find all the arrangements of 4 letters given that it must begin with a C
I divided into 2 cases:
Case 1: no repetitions, we omit a O
$4P3=4!$
Case 2: one double O. $3P2/2!$ or in other words, $(3!/2!)/2!$ to arrange the 2 O's into the 3 remaining spaces and $*3$ because of the last remaining space on the remaining letters, R, N and A
$\frac{3!/2!}{2!}*3$
So my final answer:
$4!+\frac{3!/2!}{2!}*3$
results to be something with decimals, which isn't exactly good, what have i done wrong here?
The only error is that $^3P_2$ is not $3!/2!$ but $3!/1!$; the formula for $^nP_r$ is $n!/(n-r)!$.
You could also write $^3P_2/2!$, which is the number of ways to choose two ordered positions and then forget about the order, more simply as $^3C_2$.