Question -
Answer -
Given:
P (5, r) = P (6, r –1)
By using the formula,
P (n, r) = n!/(n – r)!
P (5, r) = 5!/(5 – r)!
P (6, r-1) = 6!/(6 –(r-1))!
= 6!/(6 – r + 1)!
= 6!/(7 – r)!
So, from the question,
P (5, r) = P (6, r –1)
Substituting theobtained values in above expression we get,
5!/(5 – r)! = 6!/(7 –r)!
Upon evaluating,
(7 – r)! / (5 – r)! =6!/5!
[(7 –r) (7 – r – 1) (7 – r – 2)!] / (5 – r)! = (6 × 5!)/5!
[(7 –r) (6 – r) (5 – r)!] / (5 – r)! = 6
(7 – r) (6 – r) = 6
42 – 6r – 7r + r2 =6
42 – 6 – 13r + r2 =0
r2 –13r + 36 = 0
r2 –9r – 4r + 36 = 0
r(r – 9) – 4(r – 9) =0
(r – 9) (r – 4) = 0
r = 9 or 4
For, P (n, r): r ≤ n
∴ r =4 [for, P (5, r)]