  We consider two families of Pascal-like triangles that have all ones on the
left side and ones separated by $m-1$ zeros on the right side. The $m=1$ cases
are Pascal's triangle and the two families also coincide when $m=2$. Members of
the first family obey Pascal's recurrence everywhere inside the triangle. We
show that the $m$-th triangle can also be obtained by reversing the elements up
to and including the main diagonal in each row of the $(1/(1-x^m),x/(1-x))$
Riordan array. Properties of this family of triangles can be obtained quickly
as a result. The $(n,k)$-th entry in the $m$-th member of the second family of
triangles is the number of tilings of an $(n+k)\times1$ board that use $k$
$(1,m-1)$-fences and $n-k$ unit squares. A $(1,g)$-fence is composed of two
unit square sub-tiles separated by a gap of width $g$. We show that the entries
in the antidiagonals of these triangles are coefficients of products of powers
of two consecutive Fibonacci polynomials and give a bijective proof that these
coefficients give the number of $k$-subsets of $\{1,2,\ldots,n-m\}$ such that
no two elements of a subset differ by $m$. Other properties of the second
family of triangles are also obtained via a combinatorial approach. Finally, we
give necessary and sufficient conditions for any Pascal-like triangle (or its
row-reversed version) derived from tiling $(n\times1)$-boards to be a Riordan
array.
