Brain Teasers ? or 50 Years On........... ?

8:

2134, 2314, 3124, 3214, 3142, 3412, 4132, 4312

but that seems far too easy - is there something I've missed?

18, first digit can be 2,3, or 4 and in each case the remaining 3 digits can be arranged in 3!=6 ways  so 3x6=18. 

Hi Ian,

we are looking for even numbers only, which means that the last digit has to be a 2 or 4. I think IB has the right answer.

Innocent Bystander posted:

8:

2134, 2314, 3124, 3214, 3142, 3412, 4132, 4312

but that seems far too easy - is there something I've missed?

No

Mulberry posted:

Hi Ian,

we are looking for even numbers only, which means that the last digit has to be a 2 or 4. I think IB has the right answer.

oops - sent to the dunce chair for not reading the question carefully 

Innocent Bystander posted:

8:

2134, 2314, 3124, 3214, 3142, 3412, 4132, 4312

but that seems far too easy - is there something I've missed?

Hi IB,

As Nick says...."No". 8 is the correct answer. Well done............

..........as for the "but that seems far too easy"...........life's like that. Some things are  easy, some things aren't. Most of us, myself included, struggle to know which is which !

How many different hands of 4 cards can be dealt from a pack of 52 playing cards ?

Ian G. posted:

Mulberry posted:

Hi Ian,

we are looking for even numbers only, which means that the last digit has to be a 2 or 4. I think IB has the right answer.

oops - sent to the dunce chair for not reading the question carefully 

we've all been there Ian. It takes courage to post and even more to accept the odd goof !!

Well done !

Don Atkinson posted:

Staying with the permutation starters......

How many even numbers greater than 2,000 can be made from the integers 1, 2, 3, and 4 if each time each integer can only be used once ?

Ok, the one about "different hands of four cards" is a bit  more difficult, but i'm sure somebody will press the right buttons before too long.

Meanwhile, staying with the permutation starters, but removing the previous constraints..........

............how many four-digit numbers can be made from the integers 1, 2, 3 and 4 if each time, each integer can only be used once ? (there are no tricks, but this one does have a follow-on......)

Don Atkinson posted:

How many different hands of 4 cards can be dealt from a pack of 52 playing cards ?

There are 52 potential 1st cards, 51 potential second cards, 50 potential 3rd cards and 49 potential 4th cards giving 52!/(52-4)! possible hands, however 4! of them are identical as it doesn't matter which order the cards are dealt in. So 52!/ ( 4!  (52-4)! )   = 270725.   

Don Atkinson posted:

How many different hands of 4 cards can be dealt from a pack of 52 playing cards ?

I can deal 13 different hands from one deck of 52 playing cards, they will all be different for sure 

Matthew

Cones

Playing with a new set of naim super spfrikes I noticed that when laying them down on a flat surface the spikes formed a nice pattern with 6 spikes forming a complete circle with out any gaps. Seeing in the brochure that the spfrikes were constructed of perfectly formed cones I jumped to the conclusion that the cones must of course be made with an point angle of 60 degrees to form this circle, but wasn't quite convinced. Getting out the protractor I found that they didn't look quite like 60 degrees but it was hard to tell. So, I thought I should figure out if the magic angle was 60 degree and if not what angle these acoustic wonders were really made of.

What did I find...?

Matthew T posted:

Cones

Playing with a new set of naim super spfrikes I noticed that when laying them down on a flat surface the spikes formed a nice pattern with 6 spikes forming a complete circle with out any gaps. Seeing in the brochure that the spfrikes were constructed of perfectly formed cones I jumped to the conclusion that the cones must of course be made with an point angle of 60 degrees to form this circle, but wasn't quite convinced. Getting out the protractor I found that they didn't look quite like 60 degrees but it was hard to tell. So, I thought I should figure out if the magic angle was 60 degree and if not what angle these acoustic wonders were really made of.

What did I find...?

Sorry but I can't quite envisage what you're asking. e.g. are all the points of the cones touching each other and the flag surface, are the bottoms of the cones flat ( and if so how can the form a complete circle  surely they'd form something like a hexagon). A sketch might help us to see what you mean.

Thanks.

Ian,

Imagine a right cone laid on its side, ie in the "neutral stability" position where it can roll around its sloping surface making contact with a flat surface (table) along the cone's slant side..

Imagine a group of six such cones, apexes (top points) all meeting at a point, with each cone "just" touching the cones on either side, again along a slant side. The group of six only "just touch each other" thus forming a sort of circle.

I think Matthew is asking "what is the apex angle of such cones ?"

But i'm sure Matthew will put us both right if my interpretation is wrong !!

Don, you have it right.

Matthew

Thanks Don & Matthew, still wrestling with this one  

Ian G. posted:

Thanks Don & Matthew, still wrestling with this one  

Hi Ian,

Did my explanation help, or are you still wrestling with the general picture ?

Perhaps the picture below might help ?

Cheers, Don

Thanks Don, wrestling with the solution. Funnily enough I made some cones just like  yours - staples and all.

ian 

ok, I need to go out now and I haven't had time to check this carefully  but my (first) answer is 52.7 degrees - anyone agree or disagree ? 

And my second answer is 50.42 degrees after finding a pesky  factor of 2 in my working.  This is the total angle of the apex. 

I'm finding this a tricky problem to get my head around - I can't even persuade myself whether it should be bigger or smaller than 60 degrees! Good exercise for the grey matter. 

Ian, i'm pretty sure it's less than 60 degrees and the slope is longer than the base diameter.

Hi Ian, I'm struggling on this one too. You are not alone.

I have tried to imagine the six cones lying in a circle with their "Altitudes" all horizontal. The cones would each have angles between their base and sloping sides of 60 degrees and the Apex would be 60 degrees.

I have now tried to imagine this horizontal "ring" of cones pinned, one to each other at the point where they touch on their base diameter. I think the cones are thus free to rotate, upwards or downwards about these pivots. (but this is the bit about which I am totally uncertain ! - so I am still working on it !!!)

As they rotate, the Altitudes get longer and the Apex angles get smaller. When rotated through 90 degrees, each cone would become a cylinder ie a "cone" of infinite Altitude with an Apex angle of Zero degrees.

At some point during this rotation, the sloping sides of the cones would bcome horizontal (as opposed to the "Altitude" being horizontal at the start). I have done a calculation to estimate when this happens and the associated Apex angles.

I get 54.368 degrees.

My initial guess, (when I first read this one), was 57.3 degrees ie one radian - it seemed like an elegant sort of answer.

I get 53.54 deg

My technique was:-

Assume the pivot point remains fixed – as described in my post above this is an assumption but it looks right and seems to be consistent in the calcs............

Rotate the Apex down in small, angular increments "alpha"

Calculate a new Altitude – the Cone must elongate such that the Apex describes a vertical line.

Calculate the corresponding half-apex angle “beta” of the elongated cone.

Continue until “alpha” = “beta” (ie one sloping edge of the cone is horizontal)

two questions..............

Is my description clear ?

Is the answer right ?

I don't mind if the answers to my two questions are :-

No

Yes