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# Speakers Corner

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# How good is your maths

bumpkin
Resolved

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I was asked if I had a HDMI cable which I do but they are not all the same obviously. I use HMDI as an example,the same applies to any connector.Here comes the maths bit. If there are connectors, one big and one small, one male and one female how many cables would be needed to satisfy any situation without inter connection or adapters. Long time since I studied maths but I make it 27,how about you? Now introduce a medium size and then what is the answer. Just a fun question. There is no doubt a formula for this sort of thing if anyone can remind me.

bumpkin

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Ok, you can't connect a male to male or female to female but you can plug a male into a socket on one machine and the the end into a socket on another one. One cable, male connector both ends, no adapter.

hastelloy

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I did it empirically at 11.27am yesterday.

A to B, A to C, A to D, B to C, B to D, C to D gives 6 combinations.

If you connect A to A etc that gives another 4 but then you would need to have 8 options in order to do this:-

A1, A2, B1, B2, C1, C2, D1, D2

A1 to B1, A1 to C1, A1 to D1, A1 to A2, A1 to B2, A1 to C2, A1 to D2, B1 to C1, B1 to D1, B1 to A2, B1 to B2, B1 to C2, B1 to D2, C1 to D1, C1 to A2, C1 to B2, C1 to C2, C1 to D2, D1 to A2, D1 to B2, D1 to C2, D1 to D2, A2 to B2, A2 to C2, A2 to D2, B2 to C2, B2 to D2, C2 to D2

28 combinations

PS pedantic is my middle name!

Aitchbee

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Using the same formula as per my first early post yesterday:-

Pairing 2 from 8 options: (8x7)/(1x2) = 28.

spider9

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marvin42

I don't have a middle name, but, at the risk of repeating myself:-

You have 6 types of 'ends' - LMale, LFemale, MediumMale, MF, SM & SF. I simply called these 1,2,3,4,5 & 6 - imagine painting the numbers on each type of 'end', if that helps.

Then, as in my previous post (at 2-32pm yesterday) I made cables by putting an end 1 with an end 1, then a 1 with a 2 ...... up to a 1 with an end 6

You get 6 'pairings'.

Then you move on to connect an end 2 with a 2, then a 2 with a 3 etc but there are only 5 possible pairings this time (because you had already done 2 with 1 in the previous list, so don't need a 'double').

Moving on to connect 3 with 3, it's obvious there are now only 4 pairings.

This process continues until at connector 6 there is only one possible pairing, 6 with 6.(as all others have been done previously).

So the 'formula' is simply 6+5+4+3+2+1 =21 pairings

The method I used is based on a 'Tree Diagram', just wish it could be illustrated here!

chub_tor

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I understand some of what has been explained here but does it answer the question of how many different combinations of cable do you need in your toolbox to guarantee that you will always have one that will join two pieces of equipment that need to be connected? You may have 21 pairings but you can't make more than 10 different types of cable when you have only four different connectors. As far as I am aware the two pieces of equipment that are to receive the cables are fitted with just one type of connector each.

I think some of you are over-complicating this by having the transmitting and receiving equipment (as I see them) having more than one input or output.

Bit of a poser though isn't it? Good mental exercise.

spider9

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chub-tor

There may be confusion here, my explanation was when the question was put for small/medium and large possibilities, with male & female.

For the original question of just Large/Small and Male/Female, then the answer is 10 (Got by 4+3+2+1).

bumpkin

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By "adapter" I meant anything that changes the size or gender of whatever it is connected to. In reality that would be the easiest way of doing things the question was hypothetical. Just as well by the seem of things.

natdoor

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One way of looking at the problem of the number of cables needed is to do this in two parts. There are three male connectors of different sizes and three female. The number of ways of combining these two at a time is, using nCr, 6!/4!*2! which yields 15. However, choosing two from the total of six connectors does not include the cases where the same connector is on each end of the cable. Since there are six different connectors, six cables are needed to cater for those. Hence the total number required is 21.

Marvin42's result of 28 for the two male and two female case is correct for the number of possible combinations including the same connector at each end (8!/6!*2!). However, it should be absolutely obvious that many of these result in duplicate cables. For example, his A1 to B1 produces the same cable as A1 to B2.

finerty

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with this confusing information will you be putting it for good use bumpkin? r there plans for some kind of prject your working on

bumpkin

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finerty, I posted what I thought was a fairly straightforward question expecting a few replies giving me the answer. Seems to have got us all thinking including myself :-)