How do we get from one step to another in these factorial equations?

[homeworkhelpls]

Homework Statement

confused about this

Relevant Equations

none

this is the answer, but i don't get why k factorial multiplies the bracket, what i did was k factorial divided by the bracket

 

[BvU]

this is the answer

What's the question ?

$\$

 

[homeworkhelpls]

find n in terms of k

 

[homeworkhelpls]

What's the question ?

$\$

 

[SammyS]

View attachment 316829

It's still not clear as to just what is puzzling you about this problem - or it's solution.

Do you not understand where the following comes from?
$\displaystyle (n-k-1)! (k+1)!=(n-k)!k!$

or - Do you not understand either of the following?
$\displaystyle (k+1)!=k! \, (k+1)$

$\displaystyle (n-k)!= (n-k-1)!\,(n-k)$

or - Is it something else ?

 

[BvU]

Any context, any explanation what the notation $^nC_k$ stands for ?

$\qquad$!​

https://www.physicsforums.com/help/homeworkhelp/

$\$

 

[homeworkhelpls]

Any context, any explanation what the notation $^nC_k$ stands for ?

$\qquad$!​

https://www.physicsforums.com/help/homeworkhelp/

$\$

(n,k) used in the binomial expression formula

 

[homeworkhelpls]

i don't understand how we go from here to here only.

 

[PeroK]

i don't understand how we go from here to here only.
View attachment 316855

Can you try to prove, for example, that$$(n-k)!k! = [(n-k-1)!k!](n-k)$$

 

[homeworkhelpls]

Can you try to prove, for example, that$$(n-k)!k! = [(n-k-1)!k!](n-k)$$

no i cant prove it as i don't understand how to get to this step

 

[PeroK]

no i cant prove it as i don't understand how to get to this step

What if you take out the common factor of $k!$? Can you see that:
$$(n-k)! = (n-k-1)!(n-k)$$

 

[homeworkhelpls]

i see the k! in (n-k)! but wheres it on the right hand side?

 

[PeroK]

i see the k! in (n-k)! but wheres it on the right hand side?

Please quote the post you are referring to.

 

[SammyS]

i see the k! in (n-k)! but wheres it on the right hand side?

That's not $k!$ in $(n-k)!$.

Suppose we let $m=n-k$.

Can you see that $m!=(m-1)! \, m$ ?

 

[homeworkhelpls]

That's not $k!$ in $(n-k)!$.

Suppose we let $m=n-k$.

Can you see that $m!=(m-1)! \, m$ ?

ohhhhhh yes because m(m-1)(m-2)(m-3)! ... is part of the expression so wouldn't (n-k)! just be n! ? (-k+n)! how does that get k! ?

 

[SammyS]

ohhhhhh yes because m(m-1)(m-2)(m-3)! ... is part of the expression so wouldn't (n-k)! just be n! ? (-k+n)! how does that get k! ?

It doesn't get $k!$ .

You have (on the right hand side) :

$(n-k)!\,k! \$ to start with. Then you get the following.

$(n-k)\cdot (n-k-1)!\,k! \$

Now regroup (associative law) and then use the commutative law. to get $(n-k)$ to the end of the expression. Right?

They are using the brackets $[ \ \ ]$ simply to indicate what is common to both sides of the equation.

 

[homeworkhelpls]

(Now regroup (associative law) and then use the commutative law. to get (n−k) to the end of the expression. Right?) i don't understand this part

 

[PeroK]

(Now regroup (associative law) and then use the commutative law. to get (n−k) to the end of the expression. Right?) i don't understand this part

It's difficult to help when your answer generally is that you don't understand. I think we should begin at the beginnning here. The problem statement is that we have $n, k$ such that:
$$\binom n k = \binom n {k+1}$$And we want to find the relationship between $n$ and $k$.

As a first step, can you write down that binomial equation in terms of factorials?

 

[SammyS]

"Now regroup (associative law) and then use the commutative law. to get (n−k) to the end of the expression. Right?"
i don't understand this part

It's hard to get more basic than that, but I'll try. Don't take the following explanation as an insult to your intelligence .

Consider the product: $\displaystyle \ \ a\cdot b\cdot c$ .

According to "Order of Operations" you are to take $a$ times $b$ and then take that result times $c$ . We can write that more explicitly with parentheses as $\displaystyle \ \ (a\cdot b)\cdot c$ .

Applying the Associative Law of Multiplication, we can change the order of multiplications without altering the result. This is also referred to as "regrouping". So we now have:

$\displaystyle \ \ (a\cdot b)\cdot c=a\cdot (b\cdot c)$

Multiplication of real numbers is commutative, so now apply the Commutative Law.

$\displaystyle \ \ a\cdot (b\cdot c)=(b\cdot c)\cdot a$ .

That explains how to go from $\displaystyle (n-k)\cdot (n-k-1)!\,k! \$ to $\displaystyle \left[ (n-k-1)!\,k!\,\right ] \,(n-k)\$

 

[homeworkhelpls]

It's difficult to help when your answer generally is that you don't understand. I think we should begin at the beginnning here. The problem statement is that we have $n, k$ such that:
$$\binom n k = \binom n {k+1}$$And we want to find the relationship between $n$ and $k$.

As a first step, can you write down that binomial equation in terms of factorials?

n + k = n + k + 1

 

[PeroK]

n + k = n + k + 1

I'm not sure how to respond to that. Where and how are you studying mathematics? If are an undergraduate student you may need to talk to your tutor or a professor about being too far out of your depth.

We can only really help you with things if you have some understanding of the material.

Sorry I can't be more constructive.

 

[Steve4Physics]

n + k = n + k + 1

Hi @homeworkhelpls. Do you see any problem in posting “n + k = n + k + 1”?
Can you explain why you wrote it?

Note that there are different symbols (for the number of combinations of n items taken k at a time): $C(n,k)$ or $^nC_k$ or $_nC_k$ or $\binom n k$. You used (n,k). But they all mean the same thing.

 

[malawi_glenn]

Hint 4! = 3!×4

 

[homeworkhelpls]

Hint 4! = 3!×4

you mean 4! = 4(3)(2)(1)! ?

 

[malawi_glenn]

you mean 4! = 4(3)(2)(1)! ?

4! =4×3×2×1 and 3! = 3×2×1 thus 4! = 4×3!

 

[SammyS]

ohhhhhh yes because m(m-1)(m-2)(m-3)! ... is part of the expression so wouldn't (n-k)! just be n! ? (-k+n)! how does that get k! ?

I misread your answer in the above post.

Yes, you are correct regarding: $m!=m(m-1)(m-2)(m-3)\dots (3)(2)(1)$

So that $m!=(m)(m-1)!$.
Note that hidden in the above is that
$(m-1)(m-2)(m-3) . . . (3)(2)(1)=(m-1)((m-1)-1)((m-1)-2)((m-1)-3) . . .(3)(2)(1)$

Now, since $m!=(m)(m-1)!$ , then it follows that if $m$ is replaced by $(n-k)$ you get the following.

$(n-k)!=(n-k)\cdot((n-k)-1)!$

$\quad\quad\quad =(n-k)\cdot(n-k-1)!$