I don't get it !!!
Find an integer n satisfying {(n+5)!} divided {n! (n+4)} = 9! divide 3!
Find an integer n satisfying {(n+5)!} divided {n! (n+4)} = 9! divide 3!
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You can use the definition of the factorial function to simplify the expression. Recall, n! = n(n-1)(n-2)...(2)(1). There is a nifty trick that you can do when you have something like the left-hand side of this equation. For now, ignore the (n+4) term in the denominator -- in other words, just look at (n+5)!/n! By the definition of n!, (n+5)! = (n+5)(n+4)(n+3)(n+2)(n+1)(n)(n-1)(n-2)..… You may notice that n! is embedded within (n+5)! ---> (n+5)! = (n+5)(n+4)(n+3)(n+2)(n+1)n! Substitute this into the numerator: (n+5)(n+4)(n+3)(n+2)(n+1)n!/n! The n!'s cancel giving (n+5)(n+4)(n+3)(n+2)(n+1). But don't forget the (n+4) term in the denominator! Essentially, what you'll end up getting is something like this:
[(n+5)(n+4)(n+3)(n+2)(n+1)]/(n+4) = 9!/3!
The next part is messy. You'll need to simplify both sides of the equation and you'll end up getting a fourth-order polynomial equation that needs to be solved. Since it will be fourth-order, there will be four possible solutions for n, but only one of them will be an integer. Unless you really, really like algebra, I would suggest not doing it by hand but by using Wolfram Alpha.
Hope this helps!
[(n+5)(n+4)(n+3)(n+2)(n+1)]/(n+4) = 9!/3!
The next part is messy. You'll need to simplify both sides of the equation and you'll end up getting a fourth-order polynomial equation that needs to be solved. Since it will be fourth-order, there will be four possible solutions for n, but only one of them will be an integer. Unless you really, really like algebra, I would suggest not doing it by hand but by using Wolfram Alpha.
Hope this helps!