Math Is Fun Forum

So you want 2 quick picks, and then you want 8 more picks that are made out of the first 2 picks somehow?
Are the other 8 picks similar to the first two in that they have 1 of the 5 the same and one of the mega ball the same?
Are we looking for the number of possible combinations using these rules?

Wow, I just had a reason pop into my head as to why the above number game lessens it by one each time.
It looks easy now that i see it.
Take 25, so 24/25 times 25 is 24, so then (25 - "one")/25 times 25 = 24, so that's where the "one" comes into play.
So now if 25 is 4 times less than 100, then (4/4)(25 - "one")/25 is still the same thing, 24.
So (4x25 - 4x"one")/(4x25) times 25 = 24,  so then (100 - 4)/100 times 25 = 24.
So it is a matter of "applying" or "multiplying" the same ratio, but in a form where 100 is the denominator, like in percents,
where you just "drop" the 100 in the denominator, because % means that.

Wow, I am so happy!!!!

Oh, I am so excited by this idea!!
take the factors and remove the multipliers out and off of the x, and put the multiplier on the outside of the whole system.
wammo:
(x+2)(3x+1)(2x-3)
becomes
(x+2)3(x + 1/3)2(x-1 1/2)
so 2 times 3 is 6 so you get:
6(x+2)(x+1/3)(x-1-1/2) and then you can see the zeros right away without dividing!!
Thanks bobbundy!!

wow, bob bundy is smart, but even though he did it, i'll keep trying to learn, so here goes nothing.

I will demonstrate dividing by a guess of (x-1), i think that is called the divisor or denominator.
So x from the (x-1) goes into 6x**3, 6x**2 times, so basically 6x**3 divided by x is 6x**2 or 6x^2, same thing. (meme chose)
(meme chose is in french for same thing)
Now write down the 6x^2 above the long division shape like this:
_______________
)

Put the 6x^2 above that whole thing just like long division in 4th grade.

Now just like long division in 4th grade with just numbers, you multiply the top number you just wrote
by the number of the left of the )

So (x-1) is to the left of the )

So 6x^2 times (x-1) is 6x^3 - 6x^2.  (This step I'll name step AA11 in case you want to talk about that step.)

So write the AA11 answer below the equation (polynomial) we are starting with inside the
_________________
)

              6x**2
          ________________________________
(x-1)  )   6x**3 + 5x**2 - 17x**1 - 6x**0
             
             6x**3  - 6x**2

This is just the very beginning of dividing, it is faster on paper than by typing like this ofcourse.

The next thing to do is to compare this to a long division problem from when you were ten years old with numbers.

          8
     _____________
6  )   49
       
        48

See how they are similar to each other, 8 times 6 is 48, then you would subtract the 48 from the 49 back in school.
So that's what you do with algebra too.

I'm getting a little tired, but if you want me continue with this idea, just let me know and call it the AA22 conversation.

Okay, come back anytime and we will try harder next time...

I like small examples when I can't understand a big example.
So can you make this problem easier somehow, by cutting
down on the list size and the last number is still okay I guess.
But how about a list of just 2 numbers that can be from 0 to 9
and then a special final number that can also be 0 to 9, but the
first two numbers can't be the same right?  And also, what about
the column issue between say 5 or 10 of these 3-lists.
Can you explain that a little better, or make it smaller and easier
for me to work on?  I just think I might be able to make some
headway if we start really small, and then later work to the
actual example you are interested in.   Maybe bobbym can
handle this general case, but I frankly need smaller because
it helps me to understand what is and isn't allowed and makes
explaining the rules easier, and examples are easier to
list out in full, many times.  Sorry for the suggestion if you don't
like it, I am sorry I am so dumb.

Good Advice.  I'm a tiny bit thirsty, so I'll go do that right now...
...Yeah, national forest spring water, mm..mm..good.
I noticed something neat today on my calculator.
I'll explain by example since it is easier for me.
Take a number and compare it to 100.  Then do
something and multiply and get the next number below (one less).
Here's an example:  Take 87.5, now 87.5 is 700/8 and hence
8/7 times 87.5 is 100.  So now do (100-8/7) and get 98 and 6/7.
Multiply 87.5 by 98 and 6/7 percent and you get 86.5, one less.
This works for any number.  Take 25, 25 times 4 is 100, so
100 - 4 is 96 and 25 times 96% is 24, one less than 25.
It also works for numbers over 100.  Take 200.  200 times 1/2 is 100.
So 100 - 1/2 = 99.5 and 200 times 99.5% is 199, again 1 less than 200.
So I don't know why it works and probably won't understand anyones
formulas, but just thought i'd share that with you...

I'm down to 230 lbs.  This is pretty good for me.  Now I have to lose 5 to 15 more pounds in the next season or two.  I'm eating healthier now and a lot more veges and fruit, and limiting the deserts and processed foods to a small amount.

Why don't you just always put the numbers in numerical order (the first 5), from lowest to highest?
Wouldn't that make the number of permutations the same as the number of combinations?

maybe i could invent my own calling routine with jmps and have my very own handmade
stack-like structure in memory that tells me where to jump back to.  I could do that i suppose.
I have some 55 or so bytes of memory in the first 200 bytes of the program that i jumped over,
that i can use, but i won't do the stack growing and shinking method.  That is dangerous.  I
will just have certain memory spots for certain return locations, and i could just be careful not
to program any recursion that would mess it up, just hardcode where to jump back and then
have the subroutine grab that memory jump back offset and learn extensively how the jump
command can be used, and i probably will be able to do that in a more stable fashion than
using the stack anyway with "call" commands.