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If ` ""^(22)P_(r+1):""^(20)P_(r+2)=11:52,` find r.

Answer» We have
` ""^(22)P_(r+1):""^(20)P_(r+2)=11:52`
`rArr (22!)/({22-(r+1)}!):(20!)/({20-(r+2)}!)=11:52`
`rArr (22!)/((21-r)!):(20!)/((18-r)!)=(11)/(52)`
`rArr (22!)/((21-r)!):((18-r)!)/(20!)=(11)/(52)`
`rArr (22xx21 xx(20!))/((21-r)(20-r)(19-r)xx[(18-r)!])xx((18-r)!)/(20!)=(11)/(52)`
`rArr (22xx21)/((21-r)(20-r)(19-r))=(11)/(52)`
`rArr (19-r)(20-r)(21-r)=2xx21xx52=2xx3xx7xx4xx13`
`rArr (19-r)(20-r)(21-r)=12xx13xx14`
`rArr 19-r=12 rArr r=(19-12)=7.`
Hence, `r=7.`


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