: nPr = n-1Pr+r.n-1Pr-1
asked by guest on Jan 15, 2025 at 1:50 am
MathBot Answer:
−1Pr+n−1=n−1−n+rn+1- 1 P r + n - 1 = n - 1 - \frac{n + r}{n + 1}−1Pr+n−1=n−1−n+1n+r
Pnr=−Pr+n+rP n r = - P r + n + rPnr=−Pr+n+rSolve for PPP: P=n+rr(n+1)P = \frac{n + r}{r \left(n + 1\right)}P=r(n+1)n+rSubstitute PPP into the given expression: −1Pr+n−1=n−r(n+r)r(n+1)−1−1Pr+n−1=n−1−n+rn+1\begin{aligned}- 1 P r + n - 1 &= n - \frac{r \left(n + r\right)}{r \left(n + 1\right)} - 1 \\ - 1 P r + n - 1 &= n - 1 - \frac{n + r}{n + 1}\end{aligned}−1Pr+n−1−1Pr+n−1=n−r(n+1)r(n+r)−1=n−1−n+1n+r
