If ^nP_r = ^nP_r + 1 and ^nC_r = ^nC_r - 1, then the value of r is equal to :

Question

If ${}^n{P_r} = {}^n{P_{r + 1}}$ and ${}^n{C_r} = {}^n{C_{r - 1}}$, then the value of r is equal to :

Accepted Answer

$${}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}$$

$$ \Rightarrow (n - r) = 1$$ .....(1)

${}^n{C_r} = {}^n{C_{r - 1}}$

$$ \Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}$$

$$ \Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}$$

$$ \Rightarrow n - r + 1 = r$$

$$ \Rightarrow n + 1 = 2r$$ ..... (2)

From (1) and (2), $$  2r - 1 - r = 1 \Rightarrow r = 2$$

Solutions · 1