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you have to know combinations like 4 choose1, 4 choose2
14641 might work?
i am so lost in this one....
(5x)^4+4(5x)^3 (3)+6(5x)^2(3)^2+4(5x)(3)^3+(3)^4 if i see it right; but what do you refer to as complex?
with imaginary numbers? or simply complicated?
4C0 (5x)^4 + 4C1 (5x)^3 (3)^1 + 4C2 (5x)^2 (3)^2 + 4C3 (5x)^1 (3)^3 +4C4 (3^4)
and im finished on here, going to go to sleep
i dont know, thats what my problem says, exactly how i wrote it
^ "with imaginary numbers"? thats not even possible. They wanted the binomail expansion , they want the binomial coefficents, not just pascals triangle
then elecs answer looks to be close to what they want :)
so that long thing is the answer ? lol
thats what they mean by "complex"
4C0 (5x)^4 + 4C1 (5x)^3 (3)^1 + 4C2 (5x)^2 (3)^2 + 4C3 (5x)^1 (3)^3 +4C4 (3^4) This is the answer lol wow
I had parts of it right lol ;)
you need to multiply it out it will simplify a little bit
im still in shock that thats the long answer lol
yeah the bigger the exponent, the longer the answer
where do we get the 4c1 and stuff from? the exponent?
:) dint know if that was a fluke, or not :)
no yours above was correct 4C1=4 4C2=6
so elecengineer answer was correst right?
....i wanna say yes, but I dont have a combinatorics button on the calulator ... :/
according to an online nCr calculator, mines good lol :)
but you prolly want the 4C0, 4C1 stuff
4C0 = 1 4C1=4 4C2=6 4C3 = 4 4C4=1 they just represent numbers
nCr = n!/(n-r)!r!
wow this is way over my head