document.write( "Question 1044457: nCr:nC(r-1) = 2:3 , nC(r-2):nC(r-1) = 4:3 find n and r\r
\n" ); document.write( "\n" ); document.write( "Could someone solve it for me
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Algebra.Com's Answer #659785 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
nCr=n!/r!(n-r)!
\n" ); document.write( "nC(r-1)=n!/(r-1)!(n-r+1)!, watch the sign there.
\n" ); document.write( "nC(r-2)=n!/(r-2)(n-r+2)!
\n" ); document.write( "-------------------Divide nCr by nC(r-1)
\n" ); document.write( "n!/r!(n-r)!/n!/(r-1)!(n-r+1)! The n! cancels and you invert the denominator
\n" ); document.write( "(n-r+1)!(r-1)!/r!(n-r)!
\n" ); document.write( "(r-1)!/r!=1/r
\n" ); document.write( "(n-r+1)!/(n-r)!=n-r+1
\n" ); document.write( "The quotient for this is (n-r+1)/r=2/3
\n" ); document.write( "cross-multiply and you get 3n-3r+3=2r
\n" ); document.write( "3n+3=5r
\n" ); document.write( "------------------------
\n" ); document.write( "now do nC(r-2)/nC(r-1)
\n" ); document.write( "n!/(r-2)!(n-r+2)!/n!/(n-r+1)!
\n" ); document.write( "do the same thing with canceling the n! and inverting the denominator
\n" ); document.write( "get (r-1)!(n-r+1)!/(r-2)!(n-r+2)!
\n" ); document.write( "this is (r-1)/(n-r+2) =4/3
\n" ); document.write( "cross-multiply
\n" ); document.write( "3r-3=4n-4r+8
\n" ); document.write( "7r=4n+11
\n" ); document.write( "---------------------
\n" ); document.write( "rewrite as
\n" ); document.write( "7r-4n=11
\n" ); document.write( "5r-3n=3
\n" ); document.write( "multiply the top by 3 and the bottom by (-4)
\n" ); document.write( "21r-12n=33
\n" ); document.write( "-20r+12n=-12
\n" ); document.write( "r=21
\n" ); document.write( "substitute and n=34
\n" ); document.write( "34C21, 34C20, 34 C 19
\n" ); document.write( "The first is 927983760, the second is 1391975640, and they are in a 2:3 ratio. Really!
\n" ); document.write( "The third is 1855967520
\n" ); document.write( "The last divided by the second is 1.3333 repeat, which is 4:3
\n" ); document.write( "n is 34
\n" ); document.write( "r is 21
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