SOLUTION: please help Me These Questions 1.what will be the value of "r",if P(7,r)=60*P(7,r-3) 2.what will be the value of "n",if 3*P(2n+4,3)=2*P(n+4,4) 3.Find "n" if 3*P(n,4)=2*P(n-1,5)

Algebra ->  Permutations -> SOLUTION: please help Me These Questions 1.what will be the value of "r",if P(7,r)=60*P(7,r-3) 2.what will be the value of "n",if 3*P(2n+4,3)=2*P(n+4,4) 3.Find "n" if 3*P(n,4)=2*P(n-1,5)       Log On


   

Question 1030226: please help Me These Questions
1.what will be the value of "r",if P(7,r)=60*P(7,r-3)
2.what will be the value of "n",if 3*P(2n+4,3)=2*P(n+4,4)
3.Find "n" if 3*P(n,4)=2*P(n-1,5)
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
P(n,r) = nPr = n! / (n-r)!
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1) P(7,r) = 7Pr = 7! / (7-r)!
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60 * P(7, r-3) = 7! / (10-r)!
:
7! / (7-r)! = 60 * 7! / (10-r)!
:
1 / (7-r)! = 60 / (10-r)!
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cross multiply fractions
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(10-r)! = 60 * (7-r)!
:
(10-r)(9-r)(8-r)(7-r)(6-r) * ... = 60 * (7-r)* (6-r) * (5-r) * ...
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(10-r)(9-r)(8-r) = 60
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r can be at most 7, trial and error finds r = 5
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2 and 3 use the same method
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