Hi,
I'm working on this exercise and how it simplified the expression to get the answer confused me:
(n-1)! + (p-1)(n-1)(n-2)! = p(n-1)!
Can someone explain how the answer came to that? How do we make (n-2)! similar to, for example, 3! (3x2x1)?
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It seems that your confusion here is with the presentation of the formula; so, let me give you a slightly different description. Perhaps that description will make the notation you used make more sense.
For 
, we define 
to be the product of all natural numbers between 
and 
, inclusive: that is,
The "
" notation takes a bit of getting used to; it is basically intended to mean "and so on" or "continuing in this way". It expresses that there's a pattern at work, and that the pattern continues on in the obvious way. The first few terms are used to establish the pattern, and the final terms describe where the pattern stops.
The final (1) is just for completeness. Ending at n-(n-1) would make no difference.
The factorial is just defined as the product of the first N strictly positive integers up to N.
Now, I'm not mathematician, I'm just a lowly software engineer. But google keeps spitting out the wrong solution to this expression, and I wanted the record to be corrected.
Whenever I google for the solution to 2n!/2(n!), I'm bombarded with search results which all say the same thing: that the solution is 1.
https://www.symbolab.com/solver?or=gms&query=%5Cfrac%7B(2n!)%7D%7B(2(n!))%7D
But if we analyze even slightly, we'd easily realize this is incorrect (unless my math is just that bad).
My solution is detailed as follows:
given n! = n (n-1)(n-2)...(1) let f(n) = n! r(n) = f(2n)/(2*f(n))
If we plug in say 3 for the value of n:
r(3) = f(2*3)/(2 * f(3))
= f(6)/(2*f(3))
= 720/12
1 != 60 As you can see, for n=3 the value resolves to 60. In fact, as the value of n grows, that growth can be expressed as g(n) = 4n + 2
making the following statement true: r(n)/r(n-1) = g(n)
Being a software engineer and all, I constructed a program to validate this. Given the nature of floating point numbers and the size of the factorial results as n grows, the result has some margin of error as you exceed 15. And n > 86 are too large to represent in floating points. But i've found that plugging the same values into wolfram alpha produces the same results.
I admittedly spent way too much time looking at this, but ยฏ\_(ใ)_/ยฏ, that time cant be reclaimed now. But I wanted to share this for a couple reasons:
I wanted to demonstrate how the internet sucks and search results cant be trusted.
I wanted to be corrected if there is some flaw in my generalization.
I learned that this polynomial `4n+2` relates to something called the huckleberry rule which is a rule in chemistry for aromatic compounds, and i thought "How Cool!" and wondered if there are other patterns rooted in a more generalized form of rlike
r(n,m) = f(m*n)/m*f(n)where m is some other constant other than 2. I've tried calculating a polynomial from m=3, but I admittedly didnt get very far. so far, it feels non polynomial. but I extend it to the redditors to have fun with that question.
[EDIT]
I've learned that I'd assumed the wrong operator precedence with the factorial operator. But, I think the result is still facinating none the less.