If { }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52,t | vedclass

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(C) Given: ${ }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52$Using the formula ${ }^n P_r=\frac{n!}{(n-r)!}$,we have:$\frac{22!}{(22-(r+1))!} \div \frac{20!}{

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$7$

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(C) Given: ${ }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52$Using the formula ${ }^n P_r=\frac{n!}{(n-r)!}$,we have:$\frac{22!}{(22-(r+1))!} \div \frac{20!}{(20-(r+2))!} = \frac{11}{52}$$\Rightarrow \frac{22!}{(21-r)!} 	imes \frac{(18-r)!}{20!} = \frac{11}{52}$$\Rightarrow \frac{22 	imes 21 	imes 20!}{(21-r)(20-r)(19-r)(18-r)!} 	imes \frac{(18-r)!}{20!} = \frac{11}{52}$$\Rightarrow \frac{22 	imes 21}{(21-r)(20-r)(19-r)} = \frac{11}{52}$$\Rightarrow (21-r)(20-r)(19-r) = \frac{22 	imes 21 	imes 52}{11}$$\Rightarrow (21-r)(20-r)(19-r) = 2 	imes 21 	imes 52 = 2184$We need three consecutive integers whose product is $2184$. Testing values,$14 	imes 13 	imes 12 = 2184$.Comparing $(21-r)(20-r)(19-r) = 14 	imes 13 	imes 12$,we get $21-r = 14$,which implies $r=7$.

If ${ }^{22} P_{r+1}:{ }^{20} P_{r+2}=11: 52$,then $r=$

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