1.

If .^(22)P_(r+1) :.^(20)P_(r+2) = 11:52, find the value of r.

Answer»

Solution :`.^(22)P_(r+1): .^(20)P_(r+1) = 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)!) xx ((18-r)!)/(20!) = (11)/(52)`
`rArr (22 xx 21 xx 20!)/((21-r) xx (20-r) xx (19-r) xx(18-r)!) xx ((18-r)!)/(20!) = (11)/(52)`
`rArr (22 xx 21)/((21-r)(20-r)(19-r)) = (11)/(52)`
`rArr (21-r) (20-r) (19-r) = (22 xx 21 xx 52)/(11)`
`= 14 xx 13 xx 12`
On comparing:
`rArr 21 - r = 14`
`rArr r = 7`.


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