Given:
P (11, r) = P (12, r – 1)
By using the formula,
P (n, r) = \( \dfrac{ n!}{r!(n – r)!}\)
P (11, r) = \( \dfrac{ 11!}{r!(11 – r)!}\)
P (12, r-1) = \( \dfrac{12!}{(12 – (r-1))!}\)
= \( \dfrac{12!}{(13 – r)!}\)
So, from the question,
P (11, r) = P (12, r – 1)
Substituting the obtained values in above expression we get,
\(\dfrac{ 11!}{(11 – r)!} = \dfrac{12!}{(13 – r)!}\)
Upon evaluating,
\(\dfrac{ (13 – r)! }{ (11 – r)! }= \dfrac{12!}{11!}\)
(13 – r) (12 – r) = 12
156 – 12r – 13r + r2 = 12
156 – 12 – 25r + r2 = 0
r2 – 25r + 144 = 0
r2 – 16r – 9r + 144 = 0
r(r – 16) – 9(r – 16) = 0
(r – 9) (r – 16) = 0
r = 9 or 16
For, P (n, r): r ≤ n
r = 9 [for, P (11, r)]
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