Probability - where did I go wrong?
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Hi All - Just completed the Module 0, Week 1 Challenge Test and I am trying to understand this -
ABC&D are trying to sit in a row of 4 seats. (A) doesn't want to sit on the end seats. How many ways can they sit down?
I drew 4 seats.
For the first case, I put (A) in the second seat to the right.
Under (A), I drew three branches where B could sit.
For each B, I drew 2 possibilities where C could sit.
And carrying on, for each C, I drew 1 possibility for D to sit.
So looking at this diagram, I see: 1 A, 3 B's, 6 C's, and 6 D's
Is this not correct? Can someone help me diagram this properly? I understand that we need to diagram this twice to account for for (A) sitting in either of the two middle seats to get the actual answer, but I am stuck here!
Thanks for any help you can provide.
Best,
Will -
@inventivemonkey Hey there!
Your diagramming reasoning is correct. There are indeed 1 A, 3 B's, 6 C's, and 6 D's. Since the lowest "branching" of the diagram has six possibilities (there are 6 D's), this means that there are 6 seating possibilities if A sits in the second seat to the right.If A sits in the second seat to the left, though, nothing changes! There are still 3 places for B to sit, then 2 places for C to sit, then 1 place for D to sit from there, giving, again 1 A, 3 B's, 6 C's, and 6 D's. So the total number of ways for A, B, C, and D to sit down would be 6 + 6 = 12.
Here's another, faster way to approach it: we know that the total number of ways they can sit down equals (# places for A to sit down) x (# places for B to sit down from there) x (# places for C to sit down from there) x (# places for D to sit down from there). We know A doesn't want to sit in either end seat, so (# places for A to sit down) = 2... Now can you solve from there?
In general, things like diagramming definitely work for simpler problems where the numbers are smaller, but the second (multiplication) approach is often a lot more efficient + faster
Hope that made sense!
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@audrey Thanks! That makes perfect sense. I am realizing that this isn't "probability" at all...! I think I am confusing the concepts from another lesson & making it harder than necessary.
Much appreciated!
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@inventivemonkey No problem, glad I could help!
