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I've trying to find the answer to this but failing. I understand that if the order matters, there are 6! permutations of 6 elements, or 720. But I'm trying to figure out how many permutations there are if each of the six elements is oriented in one of two ways. Like if I have 6 arrows where each arrow can be pointing either right or left. How many permutations are there in this scenario? Any help someone can provide would be greatly appreciated. Thanks!

Jmb1978
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    How many ways are there of orienting the arrows if you keep the elements in a particular order? – Henry Nov 17 '22 at 01:21
  • 2 ways right or left (R or L). So one set (if each arrow is labelled A to F) could be: AR, BR, CL, DL, ER, FR, and another could be AL, BR, CL, DR, ER, FR. A third could be DR, FR, AL, CR, ER, BL. – Jmb1978 Nov 17 '22 at 03:00
  • There are 6 arrows, each can be oriented in 2 ways. So how many total ways can you orient the 6 arrows (for the arrows in a particular order) – Alborz Nov 17 '22 at 03:43
  • So assuming the order of the arrows always stays the same (ABCDEF), I manually wrote out all the ways the arrows could be either R or L and came up with 64 different ways: R(6),L(0) = 1 way; R(5),L(1) = 6 ways; R(4),L(2) = 15 ways; R(3),L(3) = 20 ways; R(2),L(4) = 15 ways; R(1),L(5) = 6 ways; R(0),L(6) = 1 way. Is there math that says this is right or wrong? – Jmb1978 Nov 17 '22 at 11:17
  • So $1+6+15+20+15+6+1=64$. This is $2^6$ ways since each of the six arrows can independently be oriented one of two ways. And you correctly say there are $6!=720$ ways of ordering the arrows. Can you now see the answer to your original question? – Henry Nov 17 '22 at 23:42
  • Since for each of the 720 arrangements of the 6 there are 64 possibilities for the orientation, I believe it is 720 x 64 = 46,080. – Jmb1978 Nov 17 '22 at 23:52

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