Re: [Jprogramming] Quora Problem

```It is also possible to do this combinatorially.

If allowed digits are 1,2,3, and we have six slots to use, then there are a
total of 3^6 = 729 arrangements.```
```
So we just need to count all arrangements that have 3 or more consecutive
digits that are the same.

Consider the slots as such
_ _ _ _ _ _

Use | (pipe) to denote the last slot that is in a 3-series (3 or more
consecutive digits). Important point is that we are choosing the LAST
slot that is part of a 3-or-more  series.
e.g.
_ _ _ | _ _ _
can be
1 1 1 | _ _ _
2 2 2 | _ _ _
3 3 3 | _ _ _

where the right hand of | needs to be counted.

There are 3 ways of choosing the LHS. RHS depends on whether index 4 and index
5 slots contain the same digit.
If they do then there are

3 * 2 * 2 ways.
If not then there are

3* 2 * 2 * 3 ways.

Next consider

_ _ _ _ | _ _
The first item can be any. Use 3 to denote that it canbe any three of the
digits.

3 _ _ _ | _ _
Using the previous argument we have
3 * 3 * 2 * 3 ways.

Next,
_ _ _ _ _  | _
We have 3 * 3 *3 * 2 ways.

Lastly
_ _ _ _ _ _|

We have 3 ^ 4.

Total is (3^4) + (3*3*3*2) + (3*3*2*3) + (3*2*2*3) + (3*2*2)

Subtract from the total number of possible arrangements:
729 - (3^4) + (3*3*3*2) + (3*3*2*3) + (3*2*2*3) + (3*2*2)
492

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On Mon, 4/9/18, 'Jon Hough' via Programming <programm...@jsoftware.com> wrote:

Subject: Re: [Jprogramming] Quora Problem
To: programm...@jsoftware.com
Date: Monday, April 9, 2018, 3:57 PM

Not the fastest method...

f=:
-.@:(1&e.)@:(#@:~."1)@(3&(]\))`0:@.(1&e.@:('0456789'&e.))@:":
D=: 1e5 + i. 1e6 - 1e5
+/ f"0 D
492
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On Mon, 4/9/18, Skip Cave <s...@caveconsulting.com>
wrote:

Subject: [Jprogramming] Quora Problem
To: "programm...@jsoftware.com"
<programm...@jsoftware.com>
Date: Monday, April 9, 2018, 2:51 PM

Here's a fun math challenge on Quora:

Find the number of 6-digit numbers
up of the digits 1, 2, and 3 which
have no digit recurring three or more
times, consecutively?