Given:
P (5, r) = P (6, r – 1)
By using the formula,
P (n, r) = \( \dfrac{ n!}{r!(n – r)!}\)
P (5, r) = \( \dfrac{ 5!}{r!(5 – r)!}\)
P (6, r-1) = \( \dfrac{6!}{(6 – (r-1))!}\)
= \( \dfrac{6!}{(6 – (r+1))!}\)
= \( \dfrac{6!}{(7 – r)!}\)
So, from the question,
P (5, r) = P (6, r – 1)
Substituting the obtained values in above expression we get,
\( \dfrac{5!}{(5 – r)!} = \dfrac{6!}{(7 – r)!}\)
Upon evaluating,
\(\dfrac{ (7 – r) (6 – r) (5 – r)! }{ (5 – r)!} = 6 \)
(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)]
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