Ball dropping

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ssmmss

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A single ball is dropped at the top and hits a series of pegs as it falls. At each peg, the ball can go in one of two ways (right or left) with equal probability. How many different paths are possible?

I came up with 32 paths are possible. Thoughts please.
 
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There are 6 layers and, at each layer, the ball can go in either of two directions. There are \(\displaystyle 2^6= 64\) different paths.
 
ssmmss said:
A single ball is dropped at the top (please see pic) and hits a series of pegs as it falls. At each peg, the ball can go in one of two ways (right or left) with equal probability. How many different paths are possible?
Click to expand...
I assume that the ball enters the maze at the top and drops six rows.
On each peg it has two choices: left or right.
So there are strings of (L's and R's). Each string is of length 6. How many are possible?
 
ssmmss said:
I came up with 30 paths are possible. Thoughts please.
Click to expand...

Show what you did. Then there will be something to comment on.

In general, we don't hand-out answers; next time, please demonstrate your effort or explain reasoning.

Here's a 6-min video lesson on the fundamental counting principle. Try to understand how it may be applied to your exercise.

Cheers :)
 
HallsofIvy said:
There are 6 layers and, at each layer, the ball can go in either of two directions. There are \(\displaystyle 2^6= 64\) different paths.
Click to expand...

I was looking at the base/bottom where the ball falls as it doesn't get any further than that so I didn't count the base (the one with 6 dots). So I only counted the first 5 rows. And if each peg is either L or R, then I ended up with 2^5 which is 32 paths.
 
ssmmss said:
I only [considered] the first 5 rows.
Click to expand...

Hi ssmmss:

Considering only five rows is a mistake.

Are you thinking that the falling ball, upon reaching some peg in the sixth row, will suddenly stop falling and stick to the top of that peg?

Such an interpretation is not consistent with the exercise statement.

We're told that at each peg (that the ball encounters on its way down) the ball can go left or right.

So, you may think of the peg board as mounted on the side of a wall above the floor. The ball will work its way through all six rows of pegs, and then it will continue falling toward the floor. You're counting the number of different paths possible through the peg board.

Cheers :D
 
Quaid said:
Hi ssmmss:

Considering only five rows is a mistake.

Are you thinking that the falling ball, upon reaching some peg in the sixth row, will suddenly stop falling and stick to the top of that peg?

Such an interpretation is not consistent with the exercise statement.

We're told that at each peg (that the ball encounters on its way down) the ball can go left or right.

So, you may think of the peg board as mounted on the side of a wall above the floor. The ball will work its way through all six rows of pegs, and then it will continue falling toward the floor. You're counting the number of different paths possible through the peg board.

Cheers :D
Click to expand...

Thank you all. So the end is the floor/white space and not necessarily the base of the triangle, which makes sense as to why we would get 64 paths.
 
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