# nPr- arrangements and permutations

• Jul 17th 2009, 05:21 AM
flyinhigh123
nPr- arrangements and permutations
Prove from the defintion of nPr (as in nPr = n!/(n-r)!) that
(n+1)/Pr = nPr + r.nP(r-1)
and show that nPr = (n-2)Pr + 2r. (n-2)P(r-1) + r(r-1) x (n-2)P(r-2)

if anyone could give me some techniques or some help solving these types of questions i would really appreciate it ! thankyou =D
• Jul 19th 2009, 11:26 AM
apcalculus
Quote:

Originally Posted by flyinhigh123
Prove from the defintion of nPr (as in nPr = n!/(n-r)!) that
(n+1)/Pr = nPr + r.nP(r-1)
and show that nPr = (n-2)Pr + 2r. (n-2)P(r-1) + r(r-1) x (n-2)P(r-2)

if anyone could give me some techniques or some help solving these types of questions i would really appreciate it ! thankyou =D

When trying to prove equality between two expressions we take one side of the statement and apply 'known' transformations to see if we can get to equal the other side. Here is some help with the first part. Let me know if the algebra is ok:

Right side =
$\displaystyle nPr + r nP(r-1) = \frac{n!}{(n-r)!} + r \frac{n!}{(n-(r-1))!}=$
$\displaystyle =\frac{n!}{(n-r)!} + r \frac{n!}{(n-r+1)!}=\frac{n!}{(n-r)!} + r \frac{n!}{(n-r+1)(n-r)!}=$
$\displaystyle \frac{n! (n-r+1)}{(n-r+1)(n-r)!} + r \frac{n!}{(n-r+1)(n-r)!}= \frac{n! (n-r+1)}{(n-r+1)!} + \frac{r*n!}{(n-r+1)!}=$
$\displaystyle \frac{n! (n-r+1+r)}{(n-r+1)!}= \frac{n! (n+1)}{([n+1]-r)!} = \frac{[n+1]!}{([n+1]-r)!} = (n+1)Pr =$ Left Hand Side

Good luck!!