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#21
Thank You Arc! You read the puzzle as I meant it to be read! I think it fair to say we are all right!
Thank You Arc! You read the puzzle as I meant it to be read! I think it fair to say we are all right!
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!
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
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).