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How good is your maths


bumpkin
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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.

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spider9

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bumpkin

If you had put the question in the following way, instead of the convoluted way it appeared, then any self-respecting school pupil would have got it, by simply sketching a simple tree diagram.

*I have two units to be connected. Each unit has only one socket, which may be LM or LF or SM or SF. How many cables would I need to be sure that I could always connect the two appliances?*

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marvin42

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spider9 "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"

If you have 1,2,3,4,5 and 6 how do you put 1 with 1 when you only have one 1? If you have 2 of each, then you have 12 choices - 66 options.

As previously said, half of these are duplicates (so 33).

As well as this, thinking in cables makes it worse as each cable has 2 ends so if you have 2 cables you have 4 options but only 4 ways of connecting them unless you connect the ends of a single cable together in which case it is 6 possibilities.

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bumpkin

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spider9, I would have done in hindsight, it was not my intention to confuse the issue. I have already said that I could have phrased it more clearly.

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spider9

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marvin42

I am losing the will to live, here!! Let's change some words and see if it helps.

I buy a cable with a 1 on one end and a 1 on the other..(ie LM on one end and a LM on the other end.)

Then I buy a cable with a 1 on one end and a 2 on the other.(ie LM on one end and a LF on the other end.)

Continuing the sequence as detailed in my previous posts you end up having to buy 10 cables (for the original 4 types of ends, LM,LF,SM & SF)or 21 if you want 'medium' in there as well.

There can be no question about 'how do you put 1 with 1' as you say, since we must presume the cablemakers have more than one of each type of end!!

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marvin42

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spider9

I'm talking about connectors and have a choice of 6 (3 cables). 2 from 6 is 10.

You're talking about cables - so if you have 6 possible connectors you have 15 possible cables. 2 from 15 is 105. Some connections will be duplicates but don't ask me how many.

My earlier post was wrong to divide 66 by 2 as some connections (ie 1 to 1) occur only once whilst others occur 4 times. I can't be bothered to work that out!

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marvin42

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Correction - with 6 connectors you have 21 cables and 2 from 21 is 210.

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lotvic

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I prefer WiFi and the WPS buttons

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john bunyan

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How long is a piece of string?

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Bing.alau

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When I used to do the pools back in my youth. I used to do "any eight from ten" to produce a certain number of lines. Can the mathematicians amongst us not use that system which seemed to be pretty faultless. I don't do the pools now because I do not need the money. (If you believe that you will believe anything).

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Aitchbee

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It makes one wonder, at the amount of electrical connections that the nerves in the human brain make [per second] to make mankind ['n' womankind], work out these deep mathematical problems!

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