The Binomial Theorem example problem/proof explanation needed

alexapprend

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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:
IMG_20220413_090858.jpg
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.
IMG_20220413_091309.jpg
Thanks in advance
 
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: [imath]n!=n(n-1)![/imath]

So [imath]2!=2\times 1 = 2\times 1![/imath]

Going backwards [imath] 2\times 1!=2\times 1 = 2![/imath]

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

Hope this helps
:)
 
Last edited:
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 [imath]N~\&~k[/imath] are non-negative integers such that [imath]0\le k\le N[/imath] then [imath]\dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}[/imath]
That symbol is read [imath]N\text{ choose }k[/imath].
Here are three to practice on: 1)[imath]\dbinom{10}{3}[/imath]; 2)[imath]\dbinom{5}{4}[/imath; & 3)[imath]\dbinom{7}{1}[/imath]
EXAMPLE to follow: [imath]\dbinom{15}{9}=\dfrac{15!}{9!(15-9)!}[/imath]
To see if you did it correctly SEE HERE
 
Hi alexapprend,
They're using this fact: [imath]n!=n(n-1)![/imath]

So [imath]2!=2\times 1 = 2\times 1![/imath]

Going backwards [imath] 2\times 1!=2\times 1 = 2![/imath]

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):

IMG_20220413_185139.jpg
 
First you need to practice evaluating binomial coefficients.
If [imath]N~\&~k[/imath] are non-negative integers such that [imath]0\le k\le N[/imath] then [imath]\dbinom{N}{k}=\dfrac{N!}{k!(N-k)!}[/imath]
That symbol is read [imath]N\text{ choose }k[/imath].
Here are three to practice on: 1)[imath]\dbinom{10}{3}[/imath]; 2)[imath]\dbinom{5}{4}[/imath; & 3)[imath]\dbinom{7}{1}[/imath]
EXAMPLE to follow: [imath]\dbinom{15}{9}=\dfrac{15!}{9!(15-9)!}[/imath]
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:
IMG20220413185446.jpg
Although I didn't understand 4b as the * shows, maybe you can explain that too ?

IMG_20220413_185640.jpg
IMG_20220413_185513.jpg
Some similar sort of factorial simplification questions from the exercise two before this one so you can see what I know:
IMG20220413185549.jpg
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 ?
 
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.
 
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