If we choose k elements from a set A of n elements and put it in set B ,then from the remaining n-k elements in A we can select 0 to n-k elements and put it in C. These 2 sets B,C are disjoint.
For example,A={1,2,3,4,5,6} then if B={1,6,3,5} then remaining elements in A ={2,4}.Now C can be {},{2},{3},{2,3}.These B,C form disjoint subsets.
here k ranges from 0 to n.
And l ranges from 0 to n-k.
Therefore number of ordered pairs=$\sum_{k=0}^{k=n}\left ( \binom{n}{k}*\sum_{l=0}^{l=n-k}\binom{n-k}{l}\right )$
Now,$\sum_{l=0}^{l=n-k}\binom{n-k}{j}=(1+1)^{n-k}$$=2 ^{n-k}$
=$\sum_{k=0}^{k=n}\binom{n}{k}*2^{n-k}$
=$(1+2)^{n}$
=$3^{n}$