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Question
Find r if `""^5P_r = 2^6 P_(r-1)`
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Solution
`""^5P_r = 2^6 P_(r-1)`
⇒ `(5!)/((5 - r)!) = 2 xx (6!)/((6 - r + 1)!)`
⇒ `(5!)/((5 - r)!) = (2 xx 6!)/((7 - r)!)`
⇒ `(5!)/((5 - r)!) = (2 xx 6 xx 5!)/((7 - r)(6 - r)(5 - r)!)`
⇒ 1 = `(2 xx 6)/((7 - r)(6 - r))`
⇒ (7 - r)(6 - r) = 12
⇒ 42 - 6r - 7r + r2 = 12
⇒ r2 - 13r + 30 = 0
⇒ r2 - 3r - 10r + 30 = 0
⇒ r(r - 3) - 10(r - 3) = 0
⇒ (r - 3)(r - 10) = 0
⇒ (r - 3) = 0 or (r - 10) = 0
⇒ r = 3 or r = 10
It is known that `""^nP_r = (n!)/((n - r)!) 0 ≤ r ≤ n`
∴0 ≤ r ≤ 5
Hence, r ≠ 10
∴ r = 3
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