SOLUTION: Hi, I just want to clarify something: Is there a rule for factorials that is a!-(b!c!) = ac!-bc! ? An example is 8! - (7!2!), which apparently simplifies to 8*7! - 2*7! = 6*7!

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Question 1085775: Hi, I just want to clarify something:
Is there a rule for factorials that is a!-(b!c!) = ac!-bc! ?
An example is 8! - (7!2!), which apparently simplifies to 8*7! - 2*7! = 6*7! = 30240
Could you please explain how this 'formula' works?
Thank you very much
Found 3 solutions by math_helper, natolino_2017, MathTherapy:
Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!
There is no general rule for subtraction and/or addition of factorials (none that I know of). Most of the general rules with factorials involve division and multiplication.
We CAN write:
a! - (b!c!) = b! * ( a!/b! - c! )
[ This is no different than x + yz = y( x/y + z ) where we factor out a y from each term but since x may or may not have a common factor we must show it as x/y inside the paren's. In factorials, there WILL be common factors so long as b less than a, but I think that's the most we can generally say. ]
--
For your example case: 8! - (7!2!) = 7! ( 8!/7! - 2! ) = 7!( 8 - 2 ) = 7! * 6
8!/7! simplifies to 8 because 8! / 7! = (8*7*6*…*3*2) / (7*6*…*3*2) and all but the 8 cancel.
--
OR, equivalently (more clearly?): 8! - (7!2!) = 8*7! - 2!7! = 7!(8 - 2!) = 7!(8-2) = 7! * 6
In this approach, we first rewrote 8! as 8*7! then factored out the common 7!.
—
Edit: I said there will be common factors in b! *(a!/b! - c!) "so long as b < a" but I should have said there will "ALWAYS be common factors between a! and b!" (its just if b > a you end up with 1/k where k>1).

Answer by natolino_2017(77)   (Show Source): You can put this solution on YOUR website!
what you can do is this:
8! = (8)7!
so you can factorize now as:
8(7!) -(7!)(2!) = 7! (8-2!) = 7! (8-2) = 7! (6) = (7)(6!)^2 =4.147,200.
there's not a rule for a! -(b!)(c!), because you need to know the relation between a and b.
another example is

10! - (7!(3!) = 10*9*8*(7!) - (7!3!) = 7!( 10*9*8-3!) = 7! (714) = 7! (2*3*7*17) = 3,598,560
@natolino_
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
Hi, I just want to clarify something:
Is there a rule for factorials that is a!-(b!c!) = ac!-bc! ?
An example is 8! - (7!2!), which apparently simplifies to 8*7! - 2*7! = 6*7! = 30240
Could you please explain how this 'formula' works?
Thank you very much
I don't think that there are special rules because as long as I remember doing math (factorials, specifically), each and every problem is different,

so you have to apply the existing rule, if one exists, for the particular factorial problem.



8! – (7!2!)

8(7!) – (7!2!) ----- Replacing 8! with 8(7!)

7!(8 – 2!) --------- Factoring out GCF, 7!

7!(8 – 2) ---------- Replacing 2! With 2



That's how easy this is....no COMPLEXITIES!

By the way. IGNORE the person who claims this is over 4 million. You gave the answer yet he calculates this as being over 4 million.

Ridiculous!! What are some of these people doing in here? 

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