Can you explain why the answer isn't aₙ₋₂ ​ ?

  • M0★ M1★ M2★ M3★ M4★ M5★ M6

    Module 3 Week 3 Day 11 Your Turn Part 1 Mini-Question

    I don't understand the answer explanation. Also, can you explain why the answer isn't \(a_{n-2}\)? If there are two white beads at the end, and you can't have any more whites next to it, then you have to something ending in black, aka \(a_n\). And if you want the length to be correct, it has to be \(a_{n-2}\) (since there are already two white beads).

  • ADMIN M0★ M1 M5

    @divinedolphin This is such a clever comment! I really like it! You are absolutely correct that another possible answer is

    $$ c_n = a_{n-2} $$

    It just doesn't happen to be one of the answer choices. 🤷

    Let's take a look at the given solution provided first. Our sequences are: \(a_{n},\) denoting sequences ending with a black bead, \(b_n,\) denoting sequences ending with only one white bead, and \(c_n,\) denoting sequences ending with two white beads.
     
    M3W3D11-y-part-1-mini-question-why-not-a-n-2.png
     
    In order to make a \(c_n\) sequence ending with two white beads, we can start with a \(b_{n-1}\) sequence of length \( n-1\) ending with one white bead. We cannot start with an \(a_{n-1}\) sequence, since that would give us \( \textcolor{blue}{{BW}}, \) and we cannot start with a \(c_{n-1}\) sequence, since that would give us \( \textcolor{red}{WWW},\) which isn't allowed.

     
    M3W3D11-y-part-1-mini-question-explain2.png
     

    The number of \(\textcolor{blue}{WW}\) sequences is exactly the number of \(\textcolor{blue}{BW}\) sequences of length \( n-1.\)

    $$ \boxed{ c_n = b_{n-1}} $$

    This was the given answer to the mini-question.

    Now we can iterate one step farther back. Let's think; how do we get a \(b_{n-1}\) sequence?

     
    M3W3D11-y-part-1-mini-question-explain.png
     

    If we start with a \(b_{n-2}\) sequence and add a white bead, we'll get two white beads at the end, which isn't a \(b_{n-1}\) sequence. If we start with a \(c_{n-2}\) sequence and add a white bead, we'll get three beads at the end, which isn't allowed. The only way to get a \(b_{n-1}\) sequence is to start with a \(a_{n-2}\) sequence, which ends in black, and to add a white bead at the end.

    $$ b_{n-1} = a_{n-2} $$

    And from before, we know

    $$\begin{aligned} c_{n} &= b_{n-1} \\ &= a_{n-1} \\ \end{aligned} $$

    So you are correct in thinking that \(a_{n-2}\) could be another possible answer choice. 🙂

  • M0★ M2★ M3★ M4★ M5

    @debbie
    Maybe this should be a multi-answer question?

  • ADMIN M0★ M1 M5

    @rz923 Thank you for supporting this suggestion! This would be a really clever addition to the problem. I will note it down. 🙂 🙂