The Binomial Theorem example problem/proof explanation needed

[alexapprend]

Hello, I am struggling through a HL IB textbook independently and don't understand the parts circled- I feel like I have pretty good non-calculator maths so I can manipulate algebra but I'm very lost.
From the textbook:

My notes (mostly just ended up copying from the textbook, the bits I circled show me understanding some of the steps they skipped, but whilst I can see for example 2*(2-1) does in fact equal 1!*(3-1)! I don't understand it as a general rule, will it work for all numbers? What maths theory is it using? The asterisk is where I stopped understanding. You can also see this circled above.

Thanks in advance

 

[AvgStudent]

Hello, I am struggling through a HL IB textbook independently and don't understand the parts circled- I feel like I have pretty good non-calculator maths so I can manipulate algebra but I'm very lost.
From the textbook:
View attachment 32111
My notes (mostly just ended up copying from the textbook, the bits I circled show me understanding some of the steps they skipped, but whilst I can see for example 2*(2-1) does in fact equal 1!*(3-1)! I don't understand it as a general rule, will it work for all numbers? What maths theory is it using? The asterisk is where I stopped understanding. You can also see this circled above.
View attachment 32112
Thanks in advance

Hi alexapprend,
They're using this fact: $n!=n(n-1)!$

So $2!=2\times 1 = 2\times 1!$

Going backwards $2\times 1!=2\times 1 = 2!$

You can try for higher n to see it more clearly.

Hope this helps

 

[pka]

Hello, I am struggling through a HL IB textbook independently and don't understand the parts circled- I feel like I have pretty good non-calculator maths so I can manipulate algebra but I'm very lost.
From the textbook:
View attachment 32111
My notes (mostly just ended up copying from the textbook, the bits I circled show me understanding some of the steps they skipped, but whilst I can see for example 2*(2-1) does in fact equal 1!*(3-1)! I don't understand it as a general rule, will it work for all numbers? What maths theory is it using? The asterisk is where I stopped understanding. You can also see this circled above.

First you need to practice evaluating binomial coefficients.
If $N~\&~k$ are non-negative integers such that $0\le k\le N$ then $\dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}$
That symbol is read $N\text{ choose }k$.
Here are three to practice on: 1)$\dbinom{10}{3}$; 2)[imath]\dbinom{5}{4}[/imath; & 3)[imath]\dbinom{7}{1}[/imath]
EXAMPLE to follow: $\dbinom{15}{9}=\dfrac{15!}{9!(15-9)!}$
To see if you did it correctly SEE HERE

 

[alexapprend]

Hi alexapprend,
They're using this fact: $n!=n(n-1)!$

So $2!=2\times 1 = 2\times 1!$

Going backwards $2\times 1!=2\times 1 = 2!$

You can try for higher n to see it more clearly.

Hope this helps

Thank you for responding I do know and understand this rule, you can see I used it in the example above in the bit I did understand (red box). I've shown what this would still mean for me even if I were to accept that 1*0!=1 and ignore it (obviously this is true but I just worry it's meant to be used to group or simplify things):

 

[alexapprend]

First you need to practice evaluating binomial coefficients.
If $N~\&~k$ are non-negative integers such that $0\le k\le N$ then $\dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}$
That symbol is read $N\text{ choose }k$.
Here are three to practice on: 1)$\dbinom{10}{3}$; 2)[imath]\dbinom{5}{4}[/imath; & 3)[imath]\dbinom{7}{1}[/imath]
EXAMPLE to follow: $\dbinom{15}{9}=\dfrac{15!}{9!(15-9)!}$
To see if you did it correctly SEE HERE

Thank you for the examples although a few didn't come out I do know how to do this (and I double checked on your examples) it was in the previous exercise. I've posted pictures below:

Although I didn't understand 4b as the * shows, maybe you can explain that too ?



Some similar sort of factorial simplification questions from the exercise two before this one so you can see what I know:

Thank you for taking the time to respond. I'm guessing I must be missing something obvious because both you and the other person have replied with a concept I already understand but for some reason I don't understand the working in the original question ?

 

[BigBeachBanana]

Thank you for the examples although a few didn't come out I do know how to do this (and I double checked on your examples) it was in the previous exercise. I've posted pictures below:
View attachment 32117
Although I didn't understand 4b as the * shows, maybe you can explain that too ?

View attachment 32118
View attachment 32119
Some similar sort of factorial simplification questions from the exercise two before this one so you can see what I know:
View attachment 32120
Thank you for taking the time to respond. I'm guessing I must be missing something obvious because both you and the other person have replied with a concept I already understand but for some reason I don't understand the working in the original question ?

For 4b) Consider the different scenarios when the team must have at least 1 boy and 1 girl.
You will have either:
1) three boys and one girl
2) two boys and two girls, or
3) one boy and three girls
Count the number of ways for each scenario then sum them up.