Question

Find r if ${\displaystyle {}^5{P}_r={}^6{P}_{r-1}}$

Answer 1

${\displaystyle {}^5{P}_r={}^6{P}_{r-1}}$

⇒ ${\displaystyle \frac{5!}{(5- r)!}=\frac{6!}{(6-r+1)!}}$

⇒ ${\displaystyle \frac{5!}{(5- r)!}=\frac{6\times 5!}{(7-r)!}}$

⇒ ${\displaystyle \frac{1}{(5-r)!}=\frac{6}{(7-r)(6-r)(5-r)!}}$

⇒ 1 = ${\displaystyle \frac{6}{(7-r)(6-r)}}$

⇒ (7 - r)(6 - r) = 6

⇒ 42 - 7r - 6r + r² - 6 = 0

⇒ r² - 13r + 36 = 0

⇒ r² - 4r - 9r + 36 = 0

⇒ r(r - 4 ) -9 (r - 4) = 0

⇒ r(r - 4)(r - 9) = 0

⇒ (r - 4) = 0 or (r - 9) = 0

⇒ r = 4 or r = 9

It is known that, ${\displaystyle {}^n{P}_r=\frac{n!}{(n-r)!}0\le r\le n}$

∴ 0 ≤ r ≤ 5

Hence, r ≠ 9

∴ r = 4