To go from $$(0,0)$$ to $$(6,6)$$ one must take exactly 6 Right steps (R) and 6 Up steps (U).
Total unrestricted paths $$= \binom{12}{6}=924$$.
We need paths that
(i) never pass through $$(3,3)$$ and (ii) pass through at least one of $$(2,2)$$ or $$(4,4)$$.
Let
$$A=$$ set of paths that pass through $$(2,2)$$ but avoid $$(3,3)$$,
$$B=$$ set of paths that pass through $$(4,4)$$ but avoid $$(3,3)$$.
Required number $$=|A\cup B|=|A|+|B|-|A\cap B|$$.
Step 1 : Count $$|A|$$
Paths via $$(2,2)$$:
$$ (0,0)\to(2,2): \binom{4}{2}=6 $$
$$ (2,2)\to(6,6): \binom{8}{4}=70 $$
Total through $$(2,2)$$ = $$6\times70=420$$.
Among these, paths that also pass through $$(3,3)$$:
$$ (2,2)\to(3,3): \binom{2}{1}=2 $$, $$ (3,3)\to(6,6): \binom{6}{3}=20 $$
Total = $$6\times2\times20=240$$.
Hence $$|A| = 420-240 = 180$$.
Step 2 : Count $$|B|$$
By symmetry, interchange the two segments:
$$ (0,0)\to(4,4): \binom{8}{4}=70 $$, $$ (4,4)\to(6,6): \binom{4}{2}=6 $$
Paths through $$(4,4)$$ = $$70\times6=420$$.
Those that also pass through $$(3,3)$$:
$$ (0,0)\to(3,3): \binom{6}{3}=20 $$, $$ (3,3)\to(4,4): 2 $$
Total = $$20\times2\times6=240$$.
Therefore $$|B| = 420-240 = 180$$.
Step 3 : Count $$|A\cap B|$$
These paths must go $$(0,0)\to(2,2)\to(4,4)\to(6,6)$$ while avoiding $$(3,3)$$.
Segments:
$$ (0,0)\to(2,2): 6 $$
$$ (2,2)\to(4,4): \binom{4}{2}=6 $$; among these, the ones passing through $$(3,3)$$ are
$$ (2,2)\to(3,3):2 $$ and $$ (3,3)\to(4,4):2 $$ giving $$2\times2=4$$ such paths.
So paths $$(2,2)\to(4,4)$$ avoiding $$(3,3)$$ $$=6-4=2$$.
$$ (4,4)\to(6,6): 6 $$.
Hence $$|A\cap B| = 6\times2\times6 = 72$$.
Step 4 : Inclusion-Exclusion
$$|A\cup B| = |A|+|B|-|A\cap B| = 180 + 180 - 72 = 288$$.
Thus the number of required paths is 288.
Option D which is: 288
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