It is easy to say that finding the number of ways means
finding the number of strings. If we denote the
upper move with $U$ and the
right move with $R$, we have $nU$ and $nR$ to create the
dominated strings.
At first, we assume there is no condition of
mines. Then we have $2n$ places to arrange the $U$ and $R$. We can arrange $U$ in ${2n} \choose n$ ways. $R$ can be placed in $n$ places in $n \choose n$=$1$ ways. So the answer
without restriction is $2n \choose n$
But if we take the first two terms $RU$ or $UR$, we will obviously pass through the first mine. This should be excluded. Thus,
$RU...$ $\Rightarrow$ There are $2n-2$ gaps and $(n-1)U$. So the number of arrangements=${2n-2}\choose{n-1}$. Same for $UR...$
Samely, if we take the first four terms $RRUU$ or $UURR$
[We won't take $RURU$ because it is the continuation of $RU$ counted before], we will go through the second mine. Thus,
$RRUU... \Rightarrow$ There are $2n-4$ gaps and $(n-2)U$. So the number of arrangements =${2n-4}\choose{n-2}$. Same for $UURR...$
At last, taking $(n-1)U$ and $(n-1)R$, we get: $UUU...RRR...\Rightarrow$ There are $2$ gaps and
one $U$. So the number of arrangements= $2\choose1$. Also for $RRR...UUU...$
In this way, total number of arrangments will be:
${2n \choose n}-2\times {{2n-2}\choose{n-1}}-2\times {{2n-4}\choose{n-2}}-...-2\times2$;
That we wanted 
I think I am wrong 