Advertisements
Advertisements
Question
Find the value of `"r" (""^56"C"_("r" + 2)): ""^54"P"_("r" - 1)`= 30800 : 1.
Sum
Advertisements
Solution
`""^56"P"_("r" + 2): ""^54"P"_("r" - 1)` = 30800 : 1
∴ `(""^56"P"_("r" + 2))/(""^54"P"_("r" - 1)) = 30800/1`
∴ `((56!)/((56 - "r" - 2)!))/((54!)/((54 - "r" + 1)!)) = 30800/1`
∴ `(56!)/((54 - "r" - 2)!) × ((54 - "r" + 1)!)/(54!) = 30800/1`
∴ `(56 × 55 × 54!)/((54 - "r")!) × ((55 - "r")!)/(54!) = 30800/1`
∴ `(56 × 55)/((54 - "r")!) × (55 - "r")(54 - "r")! = 30800`
∴ `55 - "r" = 30800/(56 × 55)`
∴ `55 - "r" = 30800/(3080)`
∴ 55 − r = 10
∴ r = 55 − 10
∴ r = 45
shaalaa.com
Is there an error in this question or solution?
