1. ## The Binomial Theorem/Factorials?

Im a little confused on how to do this problem.

Use the definition of [n] to show that [ 6 ] + [6] = [7] for all integers 1≤r≤6
[r ] [r-1] [r] = [ r]

The formula/definition is n!/r!(n - r!) am i correct? Help or an explanation of how to use this formula would be great! Thanks!!

2. firstly it is important to know that

$n! = n\times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1
$

for example $4! =4\times 3 \times 2 \times 1 \times = 24$

and $25! = 25 \times 24 \times \dots 3 \times 2 \times 1 \times$ = a massive number.

For calculating $\binom{n}{r} =\frac{n!}{r!(n - r!)}$

make sure you cancel as many terms as you can to make life easy

i.e. $\binom{6}{3} =\frac{6!}{3!( 6- 3!)}=\frac{6!}{3!( 3!)} =\frac{6\times 5\times 4\times 3!}{3!( 3!)} = \frac{6\times 5\times 4}{3!} = \frac{120}{6}= 20$

3. Originally Posted by pickslides
firstly it is important to know that

$n! = n\times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1
$

for example $4! =4\times 3 \times 2 \times 1 \times = 24$

and $25! = 25 \times 24 \times \dots 3 \times 2 \times 1 \times$ = a massive number.

For calculating $\binom{n}{r} =\frac{n!}{r!(n - r!)}$

make sure you cancel as many terms as you can to make life easy

i.e. $\binom{6}{3} =\frac{6!}{3!( 6- 3!)}=\frac{6!}{3!( 3!)} =\frac{6\times 5\times 4\times 3!}{3!( 3!)} = \frac{6\times 5\times 4}{3!} = \frac{120}{6}= 20$

[6] + [6] = [7] for all integers 1≤r≤6.
[r-1] [r] [r]

I think i understand what your saying, however i am still confused as how to show the proof in the above statement...
my work:
6!/(r!-1(6-r!-1)) + 6!/(r!(6-r!)) = 7!/r!(7-r!)

Is this the correct way of setting this up? If so, where do i go from here?
Im confused, but thanks a lot for your time!

4. Originally Posted by Såxon
[6] + [6] = [7] for all integers 1≤r≤6.
[r-1] [r] [r]

I think i understand what your saying, however i am still confused as how to show the proof in the above statement...
my work:
6!/(r!-1(6-r!-1)) + 6!/(r!(6-r!)) = 7!/r!(7-r!)

Is this the correct way of setting this up? If so, where do i go from here?
Im confused, but thanks a lot for your time!
Why don't you just evaluate $\left(^6_r\right)$ for all $r \in [1, 6]$?

Draw Pascal's Triangle and read off the 6th line... (the 1 on the top is the ZERO line...)

5. Hello Såxon
Originally Posted by Såxon
Im a little confused on how to do this problem.

Use the definition of [n] to show that [ 6 ] + [6] = [7] for all integers 1≤r≤6
[r ] [r-1] [r] = [ r]

The formula/definition is n!/r!(n - r!) am i correct? Help or an explanation of how to use this formula would be great! Thanks!!
I think you mean to prove that $\binom6r + \binom{6}{r-1}= \binom7r$

First, note that $r! = (r-1)!\times r$

Then

$\binom6r + \binom{6}{r-1} = \frac{6!}{r!(6-r)!}+\frac{6!}{(r-1)!(6-r+1)!}$

$= \frac{6!}{(r-1)!\times r\times(6-r)!}+\frac{6!}{(r-1)!\times(6-r)!\times (6-r+1)}$

$= \frac{6!\times (6-r+1)}{(r-1)!\times r\times(6-r)!\times(6-r+1)}+\frac{6!\times r}{(r-1)!\times(6-r)!\times (6-r+1)\times r}$

$= \frac{6!\times (6-r+1)+6!\times r}{(r-1)!\times(6-r)!\times (6-r+1)\times r}$

$= \frac{6!(6-r+1+r)}{r!(6-r+1)!}$

$= \frac{6!\times 7}{r!(7-r)!}$

$= \frac{7!}{r!(7-r)!}$

$=\binom7r$

Note that you can replace $6$ throughout this proof by $n$, and get $\binom{n}{r}+\binom{n}{r-1}= \binom{n+1}{r}$