Understanding the Simplification of Binomial Coefficients

AI Thread Summary
The discussion focuses on simplifying binomial coefficients expressed in terms of factorials, specifically proving that (n over r) equals (n-r+1)/r times (n over r-1). A participant identifies a potential error in the original equation, suggesting it should read (n-r+1)/r instead. The correct simplification involves recognizing that r(r-1)! equals r! and that (n-(r-1))! simplifies to (n-r+1)(n-r)!. Participants express confusion about the algebraic steps needed to arrive at the final result of n!/r!(n-r)!.
BMY61
Messages
3
Reaction score
0
Here is the problem i am having trouble:

Expressing the binomial coefficients in terms of factorials and simplifying algebraically show that

(n over r) = (n-2+1)/r (n over r-1)

i got that equals ((n-r+1)/r) ((n!)/((r-1)!(n-(r-1))!)) but i am trying to get that to equal n!/r!(n-r)! which would bring me back to (n over r)

i am just getting confused on what to all do in between.
hope i did no confuse anyone
 
Physics news on Phys.org
BMY61 said:
Here is the problem i am having trouble:

Expressing the binomial coefficients in terms of factorials and simplifying algebraically show that

(n over r) = (n-2+1)/r (n over r-1)

i got that equals ((n-r+1)/r) ((n!)/((r-1)!(n-(r-1))!)) but i am trying to get that to equal n!/r!(n-r)! which would bring me back to (n over r)

i am just getting confused on what to all do in between.
hope i did no confuse anyone

I believe you have an error in the statement. The right hand side should read:
(n-r+1)/r (n over r-1). However, you seem to have the next statement correct.

To get the final result, note that r(r-1)! = r!
Also (n-(r-1))!=(n-r+1)!=(n-r+1)(n-r)!
 
ahh ok, thank you
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top