Help with basic binomial coefficient

[Catbird]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

Hello. I'm currently working my way through Lang's Basic Mathematics and cannot make sense of this question:

Show that if n is a positive integer at most equal to m, then

$${m \choose n}+{m\choose n-1}={m+1 \choose n}$$

__

The answer in the textbook is given as:

1. $${m \choose n}+{m\choose n-1}={m! \over n!(m-n)!}+{m! \over (m-n+1)!(n-1)}$$

[common denominator n!(m — n + 1)!]

2. $$= {m!(m-n+1)+m!n\over n!(m-n+1)!}$$
__

I omitted the rest of the answer as I understand what follows from 2.

However I don't understand how to get such denominator from 1.
Could someone please help me?

 

[mfb]

I moved your thread to our homework section, as it is homework-like.

In (1), you can write n! as n(n-1)! and (m-n+1)! as (m-n+1)(m-n)!. Afterwards the fractions should be easy to add with the usual methods.

 

[Catbird]

Thank you for the quick reply. This makes sense!

 

[Jmm1984]

I moved your thread to our homework section, as it is homework-like.

In (1), you can write n! as n(n-1)! and (m-n+1)! as (m-n+1)(m-n)!. Afterwards the fractions should be easy to add with the usual methods.

Hello, I came here looking for the same help for the same question. Can you elaborate on your reply? Perhaps I'm too exhausted from my workplace.

Also, unrelated, I decided to self [URL='https://www.physicsforums.com/insights/self-study-basic-high-school-mathematics/']study mathematics[/URL] after many years of wanting to do so. I am rejuvenating my understanding of the basics as I only studied up to high school Calculus many moons ago. Is there a forum on this site or elsewhere for people like me? I only really see self study tips on Quora and such. Thanks.

 

[mfb]

Which part is unclear?

I am rejuvenating my understanding of the basics as I only studied up to high school Calculus many moons ago. Is there a forum on this site or elsewhere for people like me?

We have a couple of mathematics sub-forums, I'm not sure what exactly you are looking for.

 

[verty]

Jmm1984, ask a question in the academic guidance section and you'll get some recommendations how to get back into learning math.

 

[PeroK]

Hello, I came here looking for the same help for the same question. Can you elaborate on your reply? Perhaps I'm too exhausted from my workplace.

Sometimes it helps to work backwards with these things. Start from:

${m!(m-n+1)+m!n\over n!(m-n+1)!}$

And see whether you can get to:

${m! \over n!(m-n)!}+{m! \over (m-n+1)!(n-1)}$

 

[epenguin]

:

1. $${m \choose n}+{m\choose n-1}={m! \over n!(m-n)!}+{m! \over (m-n+1)!(n-1)}$$

!

 

[Jmm1984]

Sometimes it helps to work backwards with these things. Start from:

${m!(m-n+1)+m!n\over n!(m-n+1)!}$

And see whether you can get to:

${m! \over n!(m-n)!}+{m! \over (m-n+1)!(n-1)}$

I'm still not understanding something correctly after many tries. Outside of the basic concept of a factorial, I've never had worked something like this out, so perhaps there is a concept that I am not familiar with?

 

[Jmm1984]

I'm still not understanding something correctly after many tries. Outside of the basic concept of a factorial, I've never had worked something like this out, so perhaps there is a concept that I am not familiar with?

Trying to get the numerators and denominators to conform to these solutions are coming up unsuccessfull and convoluted, which makes me think I'm missing something altogether.

 

[PeroK]

I'm still not understanding something correctly after many tries. Outside of the basic concept of a factorial, I've never had worked something like this out, so perhaps there is a concept that I am not familiar with?

$(n+1)! = (n+1)(n!)$?

$\frac{ab}{ac} = \frac{b}{c}$?

$\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}$?

 

[mfb]

$(n+1)! = n(n!)$?

That should be $(n+1)! = (n+1)(n!)$

 

[PeroK]

That should be $(n+1)! = (n+1)(n!)$

Very true!

 

[verty]

Trying to get the numerators and denominators to conform to these solutions are coming up unsuccessfull and convoluted, which makes me think I'm missing something altogether.

If you have ${a \over 12} + {b \over 52}$, you have to find the lcm of 12 and 52. That is 156. Do you recall how to do that? That's what they are doing from step 1-2.

 

[Jmm1984]

If you have ${a \over 12} + {b \over 52}$, you have to find the lcm of 12 and 52. That is 156. Do you recall how to do that? That's what they are doing from step 1-2.

I'm definitely familiar with the concept of lowest common denominator.
However, I did not realize that:
(n+1)!=(n+1)(n!)

Is this a rule of factorials? Perhaps I missed this in Lang's book from which I'm studying.

 

[Jmm1984]

Very true!

I was not familiar with this. I will look into seeing if this helps me.
I suppose part of me is frustrated that I'm not understanding something at the beginning of a Basic Mathematics text. Looking more forward to reviewing more as I approach where I ended in high school with calculus, and then on from there.
I appreciate all the help you all are giving. It is very encouraging.

 

[mfb]

Is this a rule of factorials?

It can be used as their definition together with 0!=1, or it follows immediately from the definition.
n! = n(n-1)(n-2)(n-3)...1
therefore:
(n+1)! = (n+1)n(n-1)(n-2)(n-3)...1 = (n+1)n!

 

[Ray Vickson]

I'm definitely familiar with the concept of lowest common denominator.
However, I did not realize that:
(n+1)!=(n+1)(n!)

Is this a rule of factorials? Perhaps I missed this in Lang's book from which I'm studying.

It follows right away from the definition of the factorial.

More generally, for integer $0 < k < n$ we have $n! = n(n-1)(n-2) \cdots (n-k+1)\, (n-k)!,$ and that leads to another formula for the binomial:
$${n \choose k} = \frac{n (n-1) \cdots (n-k+1)}{k!} \hspace{4em}(1)$$
Eq. (1) is used to define ${n \choose k}$ for integer $k \geq 0$ but non-integer $n$. For example, we sometimes have to deal with things like ${-2 \choose k}$ or ${ 1/2 \choose k}$ for positive integer $k,$ and we do that using eq. (1).

 

[Jmm1984]

It can be used as their definition together with 0!=1, or it follows immediately from the definition.
n! = n(n-1)(n-2)(n-3)...1
therefore:
(n+1)! = (n+1)n(n-1)(n-2)(n-3)...1 = (n+1)n!

Thanks for the reply. I'll let this marinate.

 

[Jmm1984]

It follows right away from the definition of the factorial.

More generally, for integer $0 < k < n$ we have $n! = n(n-1)(n-2) \cdots (n-k+1)\, (n-k)!,$ and that leads to another formula for the binomial:
$${n \choose k} = \frac{n (n-1) \cdots (n-k+1)}{k!} \hspace{4em}(1)$$
Eq. (1) is used to define ${n \choose k}$ for integer $k \geq 0$ but non-integer $n$. For example, we sometimes have to deal with things like ${-2 \choose k}$ or ${ 1/2 \choose k}$ for positive integer $k,$ and we do that using eq. (1).

Thank you. I'll take a good look at this.