# SOLUTION: Dear math teacher, I am having difficulties with the following problem: (a)Find n; (b) n cannot equal to .... integers. (n+2)Cn = 45 (n+2)(n+1)n!/(2.1)n! = 45 and after

Algebra ->  Algebra  -> Permutations -> SOLUTION: Dear math teacher, I am having difficulties with the following problem: (a)Find n; (b) n cannot equal to .... integers. (n+2)Cn = 45 (n+2)(n+1)n!/(2.1)n! = 45 and after      Log On

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 Question 476093: Dear math teacher, I am having difficulties with the following problem: (a)Find n; (b) n cannot equal to .... integers. (n+2)Cn = 45 (n+2)(n+1)n!/(2.1)n! = 45 and after canceling n! we get (n+2)(n+1) = 90 n^2 + 3n - 88 = 0 (n+11)(n-8) = 0 n = -11 and n = 8 My answer for (a) is n = 8 only since factorial is not defined for negative numbers. For (b), our n cannot equal -11 and zero since n was canceled out above. Please let me know if my reasoning is correct. Thank you very much. I.Answer by jim_thompson5910(28715)   (Show Source): You can put this solution on YOUR website!You are correct. Good job. You can verify this by plugging in n=8 and evaluating (n+2) C n to get (n+2) C n (8+2) C 8 10 C 8 (10!)/(8!(10-8)!) (10*9*8!)/(8!2!) (10*9)/(2*1) 90/2 45 So this verifies the answer.