Q:

Converting NFA to DFA

Im trying to convert a NFA to DFA.
This is the NFA 

and this is the DFA to which i converted

Is this right?
Also when converting if i write a state as [q0,q1] is this same as [q1,q0]
edit:
Im calculating the DFA as follows
step 1:

step 2:

step 3:

Update:

A:

The states of your DFA are the $8$ subsets of $\{q_0,q_1,q_2\}$, though you may not actually need all of them. Your initial state will be the set of all initial states of the NFA; this appears to be the set $\{q_0\}$. In the NFA there are $0$-transitions from $q_0$ to both $q_0$ and $q_1$, so the DFA will have a transition $$\{q_0\}\overset{0}\longrightarrow\{q_0,q_1\}\;.$$ The NFA has no $1$-transition from $q_0$, so the $1$ transition from $\{q_0\}$ in the DFA must be to $\varnothing$, the empty set of states: $$\{q_0\}\overset{1}\longrightarrow\varnothing\;.$$
In the NFA the $0$-transitions from $q_0$ are to $q_0$ and $q_1$, and there are no $0$-transitions from $q_1$, so the $0$-transition from $\{q_0,q_1\}$ in the DFA is a loop back to $\{q_0,q_1\}$: $$\{q_0,q_1\}\overset{0}\longrightarrow\{q_0,q_1\}\;.$$ In the NFA the only $1$-transition out of $q_0$ or $q_1$ goes to $q_2$, so the DFA must have the $1$-transition $$\{q_0,q_1\}\overset{1}\longrightarrow\{q_2\}\;.$$
The NFA has no transitions from $q_2$, so in the DFA both the $0$- and the $1$-transition from $\{q_2\}$ must go to $\varnothing$: $$\{q_2\}\overset{0,1}\longrightarrow\varnothing\;.$$
Finally, the NFA obviously has no transitions that start nowhere, so in the DFA both transitions from $\varnothing$ must loop back to $\varnothing$: $$\varnothing\overset{0,1}\longrightarrow\varnothing\;.$$
The only acceptor state in the NFA is $q_2$, and the only state in the DFA that contains $q_2$ is $\{q_2\}$, so the only acceptor state in the DFA is $\{q_2\}$.
None of the other four potential states of the DFA ($\{q_1\},\{q_0,q_2\},\{q_1,q_2\},\{q_1,q_2,q_3\}$) is accessible from the initial state $\{q_0\}$, so there’s no need to work out the transitions involving them.

