If you flip a coin 9 times, you get a sequence of Heads (H) and...

a)
Since each flip has two possible outcome, {H,T}.
Then, using product rule, the total number of different sequences of heads and tails can be calculated as,
$S=2^n$
$=2^9$
$=512$
Hence, the number of different sequences of heads and tails is 512.
b) The number of different sequences of heads and tails having exactly five heads can be calculated as
$C_r^n=\frac{n!}{(n-r)!r!}$
Here, the value of r is number of exactly five heads
$C_5^9=\frac{9!}{(9-5)!5!}$
$=\frac{9\times 8\times 7\times 6\times 5!}{4!5!}$
$=\frac{9\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1\times 5!}$
$=9\times 2\times 7$
$=126$ Hence, the number of different sequences of heads and tails have exactly five heads is 126
c)
The number of different sequences having at most 2 heads means the number of heads is not more than or equal to 1.
The value of r can be represented as, $r\le 1$
$C_0^9+C_1^9=\frac{9!}{(9-0)!0!}+\frac{9!}{(9-1)!0!}+\frac{9!}{(9-1)!1!}$
$=\frac{9!}{9!}+\frac{9\times 8!}{8!}$
$=1+9=10$
Hence, the number of different sequences having at most 2 heads is 10