Say that your first draw is #x# and the second draw is #y#. If you want #x+y = 12#, you can't have #x = 12,13 or 14#. In fact, since #y# is at least one, #x+y \ge x+1 > x#

So, assume that the first draw is #x \in \{1, 2, ..., 11\}#. How many "good" values for #y# we have for each of these draws?

Well, if #x=1#, we must draw #y = 11# in order to have #x+y=12#. If #x=2#, #y# must be #10#, and so on. Since we allow replacement, we can include the case #x=y=6# as well.

So, we have #11# possible values for #x#, each yielding exactly one value for #y# in order to have #x+y=12#.

It is actually easy to enumerate all the possible ways:

#x = 1# and #y = 11#

#x = 2# and #y = 10#

#x = 3# and #y = 9#

#x = 4# and #y = 8#

#x = 5# and #y = 7#

#x = 6# and #y = 6#

#x = 7# and #y = 5#

#x = 8# and #y = 4#

#x = 9# and #y = 3#

#x = 10# and #y = 2#

#x = 11# and #y = 1#