Today's Puzzle

[mitchell65]

Thank You Arc! You read the puzzle as I meant it to be read! I think it fair to say we are all right!

 

[Gary]

There are three discs, Each on has a "7" on one side and an "8" on the other side. If thrown up radomly, how many combinations of numbers is possible. (Example "777" "888" etc)

2 * 2 * 2 = 8 combinations.

• 777
• 778
• 787
• 788
• 877
• 878
• 887
• 888

I think Kari is correct!

 

[A Guy]

My head hurts...

A Guy

 

[Arc]

My head hurts...

A Guy

 

[A Guy]

Badgers? We don't need no stinkin' badgers!!

A Guy

 

[mitchell65]

My apologies for this thread! I started it off thinking it would be a bit of fun but it got taken far more seriously than I intended, so no more Puzzles from me!

 

[A Guy]

My apologies for this thread! I started it off thinking it would be a bit of fun but it got taken far more seriously than I intended, so no more Puzzles from me!

Nah, it's all good, post another, look at the fun everyone had

A Guy

 

[Arc]

Badgers? We don't need no stinkin' badgers!!

A Guy

You have badges, now you dont want badger ..... how can I manage a badgest for you

 

[tom982]

There are three discs, Each on has a "7" on one side and an "8" on the other side. If thrown up radomly, how many combinations of numbers is possible. (Example "777" "888" etc)

If asked like this, then [Disk 1:7, Disk 2:7, Disk 3:8] is not the same combination than [Disk 1:7, Disk 2:8, Disk 3:7]. The answer is 8 combinations.

Kari is correct - in statistical theory combinations refers to permutations. Two different numbers on each of 3 disks is expressed as 2!3 (two factorial 3) which is 2 x 2 = 4 x 2 = 8.

The total number of combinations is 8.

2!3 isn't mathematically correct and would imply 2!*3, which is in fact 6. I think you mean 2^3, which is how you would calculate the number of outcomes and is 8

Some of you may have seen this before but I'll share it anyway because I think the maths behind it is fascinating. Given a set of n randomly chosen people, how large does n have to be before the probability of two people sharing a birthday becomes 0.5?

It's a well known maths problem so googling it will give you the answer in seconds, if you want to ruin it


Another favourite of mine:

Let x=y
Multiply both sides by x:
x^2=xy
Subtract y^2 from both sides:
x^2-y^2=xy-y^2
Factorise both sides:
(x+y)(x-y)=y(x-y)
Divide both sides by (x-y):
x+y=y
2y=y
2=1

Why is this not true?

Tom

 

[Arc]

Lol, that's a fallacy

From the equation, we are subject to determine the values of the variables, x and y. Isn't it? Not the values of the constants.

Taking your calculations, we get
Let x=y
Multiply both sides by x:
x^2=xy
Subtract y^2 from both sides:
x^2-y^2=xy-y^2
Factorise both sides:
(x+y)(x-y)=y(x-y)
Divide both sides by (x-y):
x+y=y
or, x=y-y
or, x=0
putting the value of x in the initial condition, we get
y=0

To determine any other values of x and y, we need at least one more equation.

Number of possible outcomes are calculated as "number of possible outcomes in a trial to the power number of trials".

Number of possible outcomes for tossing a coin thrice (as good as John's original question) =8 (2*2*2). For throwing a die thrice= 216 (6*6*6). For drawing a card 3 times (with replacement) from a pack = 140608 (52*52*52).

 

[Kari]

Arc, you are my idol! (Sorry Julian, it's Arc's turn now.) You seem to be mathematically very talented .

 

[Arc]

I am not very talented, Kari As I can say, I am just above average.
Very talented Indians live in USA

 

[clunkfish]

Well I don't know. This thread is marked as solved but three different answers have been given: 4, 8 and "2!3".

2!3 is wrong, because 2!=2 and 3=3, so 2!3=2 x 3 which is 6.

The answer depends on how you specify the question. If the sequence of numbers is important, then indeed there are 8 possible outcomes. If the sequence is not important (and throwing 3 disks randomly up in the air to see how they land rather implies that it isn't), then there are only 4 possible combinations of numbers showing.

So the answer is probably 4, could arguably be 8, but absolutely definitely is NOT 2!3....

 

[Britton30]

2|3 is notation for 2 to the 3rd power, 2x2x2, or 8.

 

[Dwarf]

The answer to this enigma is, I believe, just 4. Suppose the objects in question were just simple coins, each comprising a Head and a Tail. We therefore have the following combinations: HHH, HHT, HTT, and TTT (note that HTH and THH are identical combination-wise to HHT and, likewise, so are THT and TTH to HTT).

 

[clunkfish]

2|3 is notation for 2 to the 3rd power, 2x2x2, or 8.

No it isn't. 2! means 2 factorial, or 2 x 1. 2 with a superscript 3 means 2 to the power of 3. Here this would usually be written 2^3.

 

[Britton30]

2|3 is notation for 2 to the 3rd power, 2x2x2, or 8.

No it isn't. 2! means 2 factorial, or 2 x 1. 2 with a superscript 3 means 2 to the power of 3. Here this would usually be written 2^3.

Whatever. I didn't use an excalamtio point. it's a "|"

 

[clunkfish]

2|3 is notation for 2 to the 3rd power, 2x2x2, or 8.

No it isn't. 2! means 2 factorial, or 2 x 1. 2 with a superscript 3 means 2 to the power of 3. Here this would usually be written 2^3.

Whatever. I didn't use an excalamtio point. it's a "|"

2|3 could be used to mean 2 is a divisor of 3 - which it isn't - but is basically meaningless. I assumed it was a typo. It is never used to mean 2 raised to the power 3.

 

[Britton30]

Must be different in the UK, sorry.

 

[clunkfish]

Must be different in the UK, sorry.

I'm afraid not - it's exactly the same. The thing to do when you're wrong is learn from it, not pretend that you're correct really. That just misleads you and others, and there's already enough nonsense on the internet without adding to it.