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I'm preparing for exams and came across this question from a past paper. I've answered part a and b, but this part c I need some help to figure out please!

Let S be the set of all sequences of length 5 whose elements are letters of the English alphabet.

How many sequences from S do not start with A and end with B?

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    $\begingroup$ To clarify... "do not start with A and end with B" Do you mean "do not start with A and do start with B" or do you mean "simultaneously do not start with A and do not end with B"? $\endgroup$ – JMoravitz May 7 '19 at 18:50
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Do you want sequences that do not start with A (and end with B)? Or sequences that do not (start with A and end with B)?

In the first case, we can fix B and then note that we have $26^3 \cdot 25$ choices for the other letters, because we can choose from all $26$ letters for the middle three letters, and from $25$ letters (excluding A) for the first. So we have $26^3 \cdot 25$ possible sequences.

In the second case, we can count the number of sequences that do start with A and end with B to be $26^3$ since we fix the first and last letters and have then $26$ choices for each of the remaining three letters. The total number of possible sequences will be $26^5$, so the number of sequences that do not start with A and end with B will be $26^5 - 26^3$

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  • $\begingroup$ The question is a bit confusing itself, it would've been clearer if it said: do not start with A but end with B, or do not start with A, neither end with B. However, you did the best thing by explaining both which was really helpful. Thanks a lot! $\endgroup$ – T. Mike May 7 '19 at 19:26
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Count the total number of such sequences in $S$ (I assume letters may repeat, so there are $26^5$ different sequences). Then, count how many sequences start with $A$ and end with $B$. Since you are fixing both $A$ and $B$, there are there letters left. There are $26^3$ different sequences that do start with $A$ and end with $B$. Next, subtract.

There are

$$26^5-26^3$$

different sequences in $S$ that do not start with $A$ and end with $B$.

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  • $\begingroup$ Thanks for clarifying that with a different approach! $\endgroup$ – T. Mike May 7 '19 at 19:28
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In the Interest of Clarity:

This post is on the premise that that the statement "How many sequences from S do not start with A and end with B?" means we seek the number of sequences which...

  • ... do not start with $A$, and
  • ... do end with $B$

A similar argument for the alternative interpretation (in which you cannot start with $A$ and cannot end with $B$) can be made, though.

Also I take a more direct approach in this answer: other answers seem to have gone with the complementary approach (which is totally valid, whichever is easier for you to work with).


Hint for My Interpretation:

So, you know that each sequence has the form

$$x_1 x_2 x_3 x_4 x_5$$

where, without restriction, $x_i \in \{A,B,C,...,Y,Z\}$ for all $i$. There would be $26$ choices for each.

With the given restriction (and interpretation), you would have that $x_1 \ne A$ and $x_5 = B$.

With this in mind, how many choices are there for $x_1$? $x_2$? And so on? Since they are independent choices (no choice affects the others), you can multiply the number of possibilities together. That is to say, if you had $n_k$ choices for $x_k$, then your answer is

$$n_1 \cdot n_2 \cdot n_3 \cdot n_4 \cdot n_5$$

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  • $\begingroup$ Why $x_5 = B$ for sure? $\endgroup$ – 雨が好きな人 May 7 '19 at 18:47
  • $\begingroup$ Because it is a given restriction in the problem: "How many sequences from S do not start with A and end with B?" $\endgroup$ – Eevee Trainer May 7 '19 at 18:48
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    $\begingroup$ Oh, I see. I think there are two possible interpretations of this: we either want ‘sequences that do not (start with A and end with B)’ or ‘sequences that do not start with A (and end with B)’. $\endgroup$ – 雨が好きな人 May 7 '19 at 18:49
  • $\begingroup$ Yeah, I see what you mean... I think I'll append a note to the start of my post for clarity and ask the OP. $\endgroup$ – Eevee Trainer May 7 '19 at 18:50

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