How good is your maths

  bumpkin 20:48 16 Mar 2013

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.

  chub_tor 22:10 16 Mar 2013

OK I make it 16. If there are four variables ie Small Male (SM), Small Female (SF), Large Male (LM) and Large Female (LF) and the cable length is constant then the result is 2 to the power 4 or 16. I don't see how you can account for different lengths of cable though as you don't know how far apart the equipment is going to be so a cable that is 1 mile long could still join two pieces of equipment that are only a foot apart. Now tell me I am missing something..... Tomorrow, I am off to bed.

  bumpkin 00:42 17 Mar 2013

I knew this would cause confusion, the cable has nothing to do with it I just used that as an example. My point is how many ways are there for pairing 2 from 4 options.

  rdave13 01:01 17 Mar 2013

pairing 2 from 4 options, even Google gives up on this. :(

  Aitchbee 06:38 17 Mar 2013

Pairing 2 from 4 options: (4x3)/(1x2) = 6.

Pairing 2 from 5 options: (5x4)/(1x2) = 10.

Pairing 2 from 6 options: (6x5)/(1x2) = 15.

Pairing 2 from 100 options (100x99)/(1x2) = 4950.

  chub_tor 10:04 17 Mar 2013

This is how I saw it. Four different types Small Male. Small Female, Large Male, Large Female.

If you just take one of these you can have Sm-Sm, Sm-Sf, Sm-Lm and Sm-Lf.

Repeat that for each of the other three combinations and you have a total of 16.

  Forum Editor 10:18 17 Mar 2013

If only all data cables could be bi-directional, and all devices could have compatible connectors, how much easier our lives would be. Stephen Fry sometimes rants about cable proliferation when he's travelling, and has to take along a great bundle of them.

I'm the same - I'm so terrified of ending up with no cable to make a certain connection that I travel with a veritable bird's nest of them - I'm sure I must have given many an airport scanning operative a few uneasy moments as my bag went along the conveyor.

A long time ago I made my first trip to Moscow, and made a big thing of telling the hotel that I must have an internet connection in my room. They assured me huffily that in 'modern Russia' this was not a problem, but whe I got there I discovered that - you've guessed it - I didn't have the right laptop to modem cable. 'No problem' said the receptionist, 'we'll get someone to come to your room.' Minutes later a young, geeky-looking man arrived with a bag and proceeded to make - yes make - me a suitable cable on the spot. It took him about fifteen minutes and I was connected. I still have the cable today - and it still works perfectly.

  hastelloy 11:07 17 Mar 2013

"My point is how many ways are there for pairing 2 from 4 options."

Aitchbee is correct- A with B,C or D = 3, B with C or D = 2, C with D = 1

3+2+1=6 It's all to do with Pascal's triangle! (Google it if you really want to be confused).

  bumpkin 11:25 17 Mar 2013

27 was clearly ridiculous I need to go back to school:) 16 sounds right but I can only list 10. I can see your reasoning chub_tor but I think that would involve duplicates. Anyone know for certain?

  Aitchbee 11:45 17 Mar 2013

I sometimes like to give my local bookie a headache with the following wager [on five selections]:-

10 x ew doubles (5 x 4)/(1 x 2)

10 x ew trebles (5 x 4 x 3)/(1 x 2 x 3)

5 x ew foursomes (5 x 4 x 3 x 2)/(1 x 2 x 3 x 4)

and a 1 x ew accumulator (5 x 4 x 3 x 2 x 1)/(1 x 2 x 3 x 4 x 5)

... I'd better quit while I'm ahead ;o)

  Chronos the 2nd 11:57 17 Mar 2013

Oh dear, I did not even understand the question. Lol.

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