Multiplication and addition principle for count possibilities?/

Re: multiplication and addition principle for count possibilities???????/

In each case, you have constrained exactly two of the five coin flips.

If you had constrained all five, then of course there would be only one sequence matching the constraints.

If you had constrained four out of five, then you have one degree of freedom (one specific coin toss would be allowed to be a head or a tail), so there would be two sequences satisfying the constraints.

Can you use a similar line of reasoning for the case where exactly two of the coin flips (the first and the last) are constrained?

 

Re: multiplication and addition principle for count possibilities???????/

Yes. First, let's consider tossing N coins in a row. How many possible distinct outcomes are there? You have N stages, and each stage has 2 possible outcomes, so the total number of possible sequences is just [tex]2^N[/tex].

Now, if you are asked to count the number of sequences where the first and the last coin toss are constrained, then if you think about that a bit, that's actually equivalent to simply tossing (N-2) normal coins, plus 2 coins that only have heads (or only have tails). So then M=2 for (N-2) of the coins, and M=1 for the other 2 coins. Then you have

[tex]2^{N-2} \cdot 1^2 = 2^{N-2}[/tex]

possible outcomes.

Another way to think about it is that it's equivalent to tossing a sequence of (N-2) coins, whereas the remaining two coins were already tossed by someone else prior to the experiment.