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Question
Find r if `""^5P_r = ""^6P_(r-1)`
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Solution
`""^5P_r = ""^6P_(r-1)`
⇒ `(5!)/((5 - r)!) = (6!)/((6 - r + 1)!)`
⇒ `(5!)/((5 - r)!) = (6 xx 5!)/((7 - r )!)`
⇒ `1/((5 - r)!) = 6/((7 - r)(6 - r)(5 - r)!)`
⇒ 1 = `6/((7 - r)(6 - r))`
⇒ (7 - r)(6 - r) = 6
⇒ 42 - 7r - 6r + r2 - 6 = 0
⇒ r2 - 13r + 36 = 0
⇒ r2 - 4r - 9r + 36 = 0
⇒ r(r - 4 ) -9 (r - 4) = 0
⇒ r(r - 4)(r - 9) = 0
⇒ (r - 4) = 0 or (r - 9) = 0
⇒ r = 4 or r = 9
It is known that, `""^nP_r = (n!)/((n - r)!) 0 ≤ r ≤ n`
∴ 0 ≤ r ≤ 5
Hence, r ≠ 9
∴ r = 4
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