What Is the Origin of the Last x! in the Factorial Equation (x + y)!?

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The discussion clarifies the origin of the last "x!" in the factorial equation (x + y)! = (x + y)(x + y - 1)(x + y - 2)...(x + 1)x!. The term "x!" represents the factorial of x, which includes all integers down to 1. When calculating (x + y)!, the process continues subtracting until reaching 1, hence the inclusion of the full factorial sequence. An example with specific values illustrates that (x + y)! expands to include all terms down to x!. Understanding this concept is crucial for grasping factorial calculations.
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Why is it that (x + y)!=(x + y)(x + y - 1)(x + y - 2)...(x + 1)x!
Where did the last "x!" come from?

Thanks
 
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$$(x+y)! = (x+y)(x+y-1)(x+y-2)\ldots(x+1)(x)(x-1)\ldots(2)(1)$$
Now just rewrite the rightmost factors ##(x)(x-1)\ldots(2)(1)## as ##x!##.
 
Thank you for your quick reply. I got the form you required, but still I have the concept missing. If you don't mind explaining why did we multiply by (x+1)(x)(x−1)…(2)(1)? It seems like we get to a place where y disappears by subtraction but then again why did we add the term (x+1) and so on?
 
M. next said:
Thank you for your quick reply. I got the form you required, but still I have the concept missing. If you don't mind explaining why did we multiply by (x+1)(x)(x−1)…(2)(1)? It seems like we get to a place where y disappears by subtraction but then again why did we add the term (x+1) and so on?
The definition of the factorial of any number ##n## is ##(n)(n-1)\ldots(2)(1)##, i.e., you must keep subtracting until you get all the way down to ##1##. Therefore, when calculating ##(x+y)!##, you don't stop when you get to ##x##; you must continue all the way to ##1##.
 
Try it with a concrete example if it's still unclear. For example, if ##x = 3## and ##y = 4##, then ##x+y = 7##, and ##(x+y)! = 7! = (7)(6)(5)(4)(3)(2)(1) = (7)(6)(5)(4)3!##.
 
Thank you a lot!
 
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