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Expansion of (x^2 - y^2)

If 11Pr-1:1...

Question

If $^{11} P_{r - 1} :^{11} P_{r - 2} = 14 : 3$ then find $r$ .

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Solution

$^{11} P_{r - 1} :^{11} P_{r - 2} = 14 : 3$

$⟹ \frac{11!}{(11 - r + 1)!} : \frac{11!}{(11 - r + 2)!} = 14 : 3$

$⟹ \frac{1}{(11 - r + 1)!} : \frac{1}{(11 - r + 2) (11 - r + 1)!} = 14 : 3$

$⟹ 1 : \frac{1}{(11 - r + 2)} = 14 : 3$

$⟹ 11 - r + 2 = \frac{14}{3}$

$⟹ 33 - 3 r + 6 = 14$

$⟹ 3 r = 33 + 6 - 14 = 25$

$⟹ r = \frac{25}{3} = N o i n t e g e r$

There is no such value of $r$

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