P (15, r – 1) = \(\dfrac{ 15! }{ (15 – r + 1)!}\)
= \( \dfrac{15!} { (16 – r)!}\)
P (16, r – 2) = \(\dfrac{ 16!}{(16 – r + 2)!}\)
= \( \dfrac{16!}{(18 – r)!}\)
So, from the question,
\(\dfrac{ P(15, r – 1) }{ P(16, r – 2)} = \dfrac{3}{4}\)
Substituting the obtained values in above expression we get,
\( \dfrac{15! }{ (16 – r)! } \dfrac{16!}{(18 – r)!} =\dfrac{ 3}{4}\)
(18 – r) (17 – r) = \( \dfrac{ 3}{4}\) × 16
(18 – r) (17 – r) = 12
306 – 18r – 17r + r2 = 12
306 – 12 – 35r + r2 = 0
r2 – 35r + 294 = 0
r2 – 21r – 14r + 294 = 0
r(r – 21) – 14(r – 21) = 0
(r – 14) (r – 21) = 0
r = 14 or 21
For, P(n, r): r ≤ n
r = 14 [for, P(15, r – 1)]
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