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So the problem is: Consider tossing a coin, with outcomes, $H$&$T$, and $P(H) = a$, independently for $5$ times. Each time, if you get a head, you pick up $X = x_0$ green balls and if you get a tail, you pick up $Y = y_0$ red ball. You then put the balls in a row, following the order of obtaining them, to provide a pattern. You find out you will totally have $M$ different patterns. Which of the following statement is correct?

$(A)$ $ M$ depends on $a$ (is a function of $a$).

$(B)$ You would have a larger $M$ if $x_0$ > $y_0$.

$(C)$ $M$ does not change with the values of $(x_0,y_0)$.

So the answer is $C$. But why isn't A correct? if $P(H) = 0$, wouldn't $M$ always be 1 then? If $P(H) = 0.5$, then $M$ is absolutely not $1$. So I think $M$ depends on $a$?

And for $C$,the solution says $M$ will always be $2^5$, why is that?

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    Well, it depends on $a$ at the extremes. if $P(H)=0$ or $1$ then you only get $1$ possible pattern, not so for any other value of $a$. for any other $a$, then indeed all $32$ outcomes are possible (though of course some have lower probability than others). – 2017-02-07

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To expand on what lulu has already said, and to answer your second question:

You are correct that for $P(H) = 0, M$ is $1$. And when $P(H) = 1$, $M$ is $1$ as well. I suspect the questioner overlooked this possibility, or else they would have been more circumspect in their wording. There is also nothing that prevents $x_0$ or $y_0$ from being $0$, and if exactly one of those are $0$ (and $0 < a < 1$), then $M = 6$. If both are $0$, then $M = 1$ again. So what happens when $0 < a < 1, x_0 > 0$ and $y_0 > 0$?

Let G represent $x_0$ green balls in a row. Let R represent $y_0$ red balls in a row.

The first flip can be either heads or tails. If it is heads, the pattern is G. If it is heads, the pattern is R. The value of $a$ determines only how likely these two patterns are to appear, not what patterns are available.

The 2nd flip will append another G or R to the end of the pattern of the first flip. So the possible patterns are GG, GR, RG, RR. Again, as long as $a$ is not $0$ or $1$, all patterns are possible, even if some are more likely than others.

And so on. the 3rd flip gives GGG, GGR, GRG, GRR, RGG, RGR, RRG, RRR as its possibilities, exactly twice as many as the 2nd flip had, because we are adding two possible results to the end of the previous pattern. So the Total counts are:

$$\begin{array}{c|c} \text{Flips} & M \\ \hline 1 & 2^1 = 2\\ 2 & 2^2 = 4\\ 3 & 2^3 = 8\\ 4 & 2^4 = 16\\ 5 & 2^5 = 32\end{array} $$