How to Start a Problem I'm Struggling With

[lema21]

Homework Statement

Prove following theorem by expressing all the binomial coefficients in terms of factorials and then simplifying algebraically: For any positive integer n and r=1, 2, …, n-1; (n over r) = (n-1 over r) + (n-1 over r-1).

Relevant Equations

None

I really don't know what to do for this problem. I looked at similar threads but couldn't seem to grasp the idea of it. I would like help on how to start.

 

[Delta2]

Do you know this formula ?$$(n \text{ over } r)=\frac{n!}{r!(n-r)!}$$

 

[lema21]

Yes, I just looked at a youtube video and I saw that in order to prove this type of problem I would have to apply the formula that you stated to (n-1 over r) + (n-1 over r-1). And I did that and got [ (n-1)! / (r!(n-1-r)!) ] + [ (n-1)! / ((r-1)!(n-r)!) ] but ended getting an extremely long expression which is wrong (I think) because aren't I supposed to get (n over r) = (n over r) as my answer?

 

[Delta2]

Yes after you work algebraically that sum of the two fractions of factorials you are supposed to get $\frac{n!}{r!(n-r)!}$.
To start working towards the right direction, multiply the first fraction by $\frac{n-r}{n-r}$ and the second fraction by $\frac{r}{r}$ and tell me what you get.

 

[Delta2]

have in mind that $r!=r(r-1)!$ and $(n-r)!=(n-r)(n-1-r)!$.

 

[lema21]

Yes after you work algebraically that sum of the two fractions of factorials you are supposed to get $\frac{n!}{r!(n-r)!}$.
To start working towards the right direction, multiply the first fraction by $\frac{n-r}{n-r}$ and the second fraction by $\frac{r}{r}$ and tell me what you get.

I got 2(n+1)! / (r-1)! (n-1-r!)

 

[Delta2]

I got 2(n+1)! / (r-1)! (n-1-r!)

Hmm can't tell where your mistake is unless you post the in between steps, but I think you did something terribly wrong because you should get in the common denominator of the two fractions $r!(n-r)!$ not $(r-1)!(n-1-r)!$.

 

[Delta2]

I can do the second fraction for you which is abit more easy, it is $$\frac{r}{r}\frac{(n-1)!}{(r-1)!(n-r)!}=\frac{r(n-1)!}{r(r-1)!(n-r)!}=\frac{r(n-1)!}{r!(n-r)!}$$

 

[lema21]

Oh, my previous reply was my simplified answer after adding both fractions. I don't see how you simplified the denominator from the middle to the last part.

 

[lema21]

Oh, my previous reply was my simplified answer after adding both fractions. I don't see how you simplified the denominator from the middle to the last part.

Oh nevermind... I get it.

 

[lema21]

For the first fraction I got (n-r)(n-1)! / r!(n-r)!

 

[Delta2]

ok so now the denominators of the two fractions are equal and what we want them to be (remember we want to prove that the sum is equal to $\frac{n!}{r!(n-r)!}$), we have to work the numerator of the sum of the two fractions. What do you get there?

 

[lema21]

I got (n-r)(n-1)! + r(n-1)!

 

[Delta2]

You are doing fine, how can you expand the first term using the distributive property?

 

[Delta2]

if you are getting confused by the factorial (n-1)! put z=(n-1)! and right the numerator as (n-r)z+rz :D.

 

[lema21]

Okay so I factored out the (n-1)! and got (n-1)! [(n-r)+r] = n(n-1)!

 

[Delta2]

Great! so what is n(n-1)! equal to?

 

[lema21]

n!

 

[lema21]

Thank you so much for your help :) I really do appreciate it.

 

[Delta2]

No problem I am glad that I helped a smart and polite student :)