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February 11[edit]

Permutations with fixed items[edit]

Hi,

As part of a data conversion exercise I've had to quantify and label working patterns, the naming convention I'm using looks like this:

1001100

A character is either a 1 or a 0 (working or not working), and the string always starts with a 1, as the working pattern can start on any of the 7 days (i.e. 1000000 is working one day, with 6 days off, but can start on any day.) There is also always at least one 0 (no one can work every single day). From this I've been trying to work out how many different permutations there. From our data we have 34 different permutations, but I'd like to know if this is all of them or if we are missing any.

I think the calculation is something like 7! / 5! (42), but I'm having difficulty getting my head around permutation calculations.

Is this correct or have I missed something?

Thanks 85.159.128.109 (talk) 16:59, 11 February 2016 (UTC)[reply]

If I understand you:

1) Each string contains 7 characters.

2) The first character is always "1".

3) The remaining characters can be "0" or "1", but they can't all be "1".

If I have this correctly, we can ignore the first character, since it's always the same, and the rest give us 2⁶ combos, or 64 possibilities. Discounting the "all ones" possibility leaves us with 63. StuRat (talk) 17:06, 11 February 2016 (UTC)[reply]

I agree with StuRat, however, since you wrote "the working pattern can start on any of the 7 days". I wonder if you meant that a pattern of 1001000 starting on Monday is the same as a pattern of 1000100 starting on Thursday. If they are considered the same pattern, then there are only 19 minus 1 = 18 distinct patterns. Dhrm77 (talk) 17:46, 11 February 2016 (UTC)[reply]

The appropriate Wikipedia article is Necklace (combinatorics), see also http://mathworld.wolfram.com/Necklace.html and http://oeis.org/A000031 with the caveat that *both* single color necklaces are excluded. (so 20-2=18)

1000000 (All patterns for working one day are the same)
1100000 (This is equivalent to 1000001)
1010000
1001000
1110000
1101000
1100100
1100010
1010100
1111000
1110100
1110010
1101100
1101010
1111100
1111010
1110110
1111110

Naraht (talk) 19:03, 11 February 2016 (UTC)[reply]

Hi all,

Thanks for all the great answers, I think Naraht has hit it on the head. 85.159.128.109 (talk) 10:30, 12 February 2016 (UTC)[reply]

Only based on the way that Dhrm77 (talk · contribs) was thinking about it. :)Naraht (talk) 15:42, 12 February 2016 (UTC)[reply]