document.write( "Question 1110201: Prove (n+1)p(r) = (n)p(r) + 4.(n)p(r-1)
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Algebra.Com's Answer #725206 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Interesting problem.... \n" ); document.write( "But you didn't state the equation correctly. The equation you show has many counterexamples. \n" ); document.write( "For n=6 and r=3, the formula says \n" ); document.write( "7P3 = 6P3+4*6P2. But \n" ); document.write( "7P3 = 7*6*5 = 210 \n" ); document.write( "6P3 = 6*5*4 = 120 \n" ); document.write( "6P2 = 6*5 = 30 \n" ); document.write( "and \n" ); document.write( "210 = 120 + 4*30 \n" ); document.write( "is not true. \n" ); document.write( "The interesting problem is when the formula is stated correctly: \n" ); document.write( "(n+1)Pr = nPr + r*nP(r-1) \n" ); document.write( "This is easy to prove algebraically, although the nomenclature is ugly.... \n" ); document.write( "(n+1)Pr = (n+1)(n)(n-1)(n-2)...(n-(r-2)) = (n+1)(n)(n-1)(n-2)...(n-r+2) \n" ); document.write( "nPr = (n)(n-1)(n-2)...(n-(r-2))(n-(r-1)) = (n)(n-1)(n-2)...(n-r+2)(n-r+1) \n" ); document.write( "nP(r-1) = (n)(n-1)(n-2)...(n-(r-2)) = (n)(n-1)(n-2)...(n-r+2) \n" ); document.write( "Then \n" ); document.write( "nPr + r*nP(r-1) = \n" ); document.write( "[(n)(n-1)(n-2)...(n-r+2)](n-r+1) + r[(n)(n-1)(n-2)...(n-r+2)] = \n" ); document.write( "[(n)(n-1)(n-2)...(n-r+2)][(n-r+1)+r] = \n" ); document.write( "[(n)(n-1)(n-2)...(n-r+2)](n+1)= \n" ); document.write( "(n+1)(n)(n-1)(n-2)...(n-r+2) \n" ); document.write( "(n+1)Pr \n" ); document.write( " |