document.write( "Question 1044457: nCr:nC(r-1) = 2:3 , nC(r-2):nC(r-1) = 4:3 find n and r\r
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document.write( "Could someone solve it for me \n" );
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Algebra.Com's Answer #659785 by Boreal(15235)![]() ![]() 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 \n" ); document.write( " |