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OpenStudy (anonymous):
P(2n+4,3) = 2/3 P(n+4,4)
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OpenStudy (triciaal):
what is 4,3?
OpenStudy (anonymous):
btw apperently the answer is 13? oh and we are dealing with P(n,r) stuff or statististics so n=2n+4 and r=3
ganeshie8 (ganeshie8):
(2n+4)(2n+3)(2n+2) = 2/3(n+4)(n+3)(n+2)(n+1)
ganeshie8 (ganeshie8):
cancel cmmon factors and solve n ?
OpenStudy (triciaal):
i had n = 10
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OpenStudy (anonymous):
ok give me some time :P gotta solve it to see if i get it
OpenStudy (triciaal):
@ganeshie8 would it work the same as 4C3 and use factorial? then equate and solve for n?
ganeshie8 (ganeshie8):
for 4C3 we need to divide by r! also so thats an extra i think
ganeshie8 (ganeshie8):
but permutation is much simpler :
4P3 = 4x3x2
ganeshie8 (ganeshie8):
nPr = n(n-1)(n-2)...(n-r+1)
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ganeshie8 (ganeshie8):
nCr = [ n(n-1)(n-2)...(n-r+1) ] /r!
ganeshie8 (ganeshie8):
so i think combination may give us a messy equation
OpenStudy (triciaal):
thanks
OpenStudy (anonymous):
are we supposed to get 22n^3+86n^2+74n+78=0 at the end?
ganeshie8 (ganeshie8):
(2n+4)(2n+3)(2n+2) = 2/3(n+4)(n+3)(n+2)(n+1)
2(n+2)(2n+3)2(n+1) = 2/3(n+4)(n+3)(n+2)(n+1)
4(2n+3) = 2/3(n+4)(n+3)
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