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)\"\" \"About 
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\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
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