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EXERCISE 3.3 · Q58

Q.Find r, if 12Pr−2:11Pr−1=3:14{}^{12}P_{r-2} : {}^{11}P_{r-1} = 3:14

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12Pr−2=12!(14−r)!{}^{12}P_{r-2} = \dfrac{12!}{(14-r)!} and 11Pr−1=11!(12−r)!{}^{11}P_{r-1} = \dfrac{11!}{(12-r)!}. Their ratio is 12!(14−r)!×(12−r)!11!=12×(12−r)!(14−r)!=12(14−r)(13−r)\dfrac{12!}{(14-r)!}\times\dfrac{(12-r)!}{11!} = \dfrac{12\times(12-r)!}{(14-r)!} = \dfrac{12}{(14-r)(13-r)} (since (14−r)!=(14−r)(13−r)(12−r)!(14-r)!=(14-r)(13-r)(12-r)!). Setting 12(14−r)(13−r)=314\dfrac{12}{(14-r)(13-r)}=\dfrac{3}{14}: cross-multiplying, 12×14=3(14−r)(13−r)⇒168=3(14−r)(13−r)⇒(14−r)(13−r)=5612\times14 = 3(14-r)(13-r) \Rightarrow 168=3(14-r)(13-r) \Rightarrow (14-r)(13-r)=56. Since 56=8×756=8\times7, take 14−r=8, 13−r=714-r=8,\ 13-r=7, giving r=6r=6 (both consistent). Check: 12P4=11880, 11P5=55440{}^{12}P_4=11880,\ {}^{11}P_5=55440, ratio 11880:55440=3:1411880:55440 = 3:14. ✓

✓Final answer

r = 6

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