If P (11, r) = P (12, r – 1), find r.

1.	If P (11, r) = P (12, r – 1), find r.
Answer» Given as P (11, r) = P (12, r – 1) On using the formula, P (n, r) = n!/(n – r)! P (11, r) = 11!/(11 – r)! P (12, r-1) = 12!/(12 – (r-1))! = 12!/(12 – r + 1)! = 12!/(13 – r)! Therefore, from the question, P (11, r) = P (12, r – 1) On substituting the obtained values in above expression we get, 11!/(11 – r)! = 12!/(13 – r)! Now, upon evaluating, (13 – r)! / (11 – r)! = 12!/11! [(13 – r) (13 – r – 1) (13 – r – 2)!] / (11 – r)! = (12×11!)/11! [(13 – r) (12 – r) (11 -r)!] / (11 – r)! = 12 (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)]

Answer»

Given as

P (11, r) = P (12, r – 1)

On using the formula,

P (n, r) = n!/(n – r)!

P (11, r) = 11!/(11 – r)!

P (12, r-1) = 12!/(12 – (r-1))!

= 12!/(12 – r + 1)!

= 12!/(13 – r)!

Therefore, from the question,

P (11, r) = P (12, r – 1)

On substituting the obtained values in above expression we get,

11!/(11 – r)! = 12!/(13 – r)!

Now, upon evaluating,

(13 – r)! / (11 – r)! = 12!/11!

[(13 – r) (13 – r – 1) (13 – r – 2)!] / (11 – r)! = (12×11!)/11!

[(13 – r) (12 – r) (11 -r)!] / (11 – r)! = 12

(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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