Factorials (combinations, pascal's triangle)

burgerandcheese

burgerandcheese

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85
combinations.jpg

Hi

From the picture above, regarding the numerator only, how did they get n(n - 1)....(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 = n!
?

It just doesn't make sense to me. I tried it with real numbers too. Let n = 7, r = 3

7(7 - 1)(7- 1 - 1)(7 - [3 - 1])(7 - 3)(7 - 3 - 1])....*2*1
= 7(7 - 1)(7 - 2)(7 - 2)(7 - 3)(7 - 4)...*2*1

(7 - 2) appears twice, so it's not 7!, it's 7!*(7 - 2)

Maybe I missed something? Maybe it's supposed to be read in another way? I fully understand how we get from n! / [(n - r)!*r!) to [n(n - 1)(n - 2)...(n - [r - 1)] / r! , but this picture tells me how to get from the latter to the former, and I don't understand it!

Source:
http://gcecompilation.com/wp-content/uploads/2017/Pure%20Mathematics%20-1.pdf

page 136
 
burgerandcheese said:
View attachment 10817

Hi

From the picture above, regarding the numerator only, how did they get n(n - 1)....(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 = n!
?

It just doesn't make sense to me. I tried it with real numbers too. Let n = 7, r = 3

7(7 - 1)(7- 1 - 1)(7 - [3 - 1])(7 - 3)(7 - 3 - 1])....*2*1
= 7(7 - 1)(7 - 2)(7 - 2)(7 - 3)(7 - 4)...*2*1

(7 - 2) appears twice, so it's not 7!, it's 7!*(7 - 2)

Maybe I missed something? Maybe it's supposed to be read in another way? I fully understand how we get from n! / [(n - r)!*r!) to [n(n - 1)(n - 2)...(n - [r - 1)] / r! , but this picture tells me how to get from the latter to the former, and I don't understand it!

Source:
http://gcecompilation.com/wp-content/uploads/2017/Pure%20Mathematics%20-1.pdf

page 136
Click to expand...
Weird book.

You know that \(\displaystyle \displaystyle \Sigma\) means add them up, right?

Well \(\displaystyle \displaystyle \Pi\) is analogous but means to multiply instead of add.

All they are saying is this

\(\displaystyle \dbinom{7}{2} = \dfrac{7!}{2! * (7 - 2)!} = \dfrac{7 * 6 * 5!}{2! * 5!} = \dfrac{7 * 6}{2}.\)

More generally, \(\displaystyle \text {if } 0 \le r < n,\)

\(\displaystyle \dbinom{n}{r} = \dfrac{n!}{r! * (n - r)!} = \dfrac{\displaystyle \prod_{j=2}^n j}{\displaystyle r! * \left ( \prod_{j=2}^{n-r} j \right)} = \)

\(\displaystyle \dfrac{\displaystyle \left ( \prod_{j=2}^{n-r} j \right) * \left ( \prod_{j=(n-r)+1}^n j \right )} {\displaystyle \left ( \prod_{j=2}^{n-r} j \right) * r!} = \dfrac{\displaystyle \prod_{j=(n-r)+1}^n j}{r!}.\)

Clear now?

Let's do one more example.

\(\displaystyle \dbinom{12}{4} = \dfrac{12!}{4! * (12 - 4)!} = \dfrac{12 * 11 * 10 * 9 * 8!}{4! * 8!} = \dfrac{12 * 11 * 10 * 9}{4 * 3 * 2}.\)
 
Last edited:
burgerandcheese said:
From the picture above, regarding the numerator only, how did they get n(n - 1)....(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 = n!
?

It just doesn't make sense to me. I tried it with real numbers too. Let n = 7, r = 3

7(7 - 1)(7- 1 - 1)(7 - [3 - 1])(7 - 3)(7 - 3 - 1])....*2*1
= 7(7 - 1)(7 - 2)(7 - 2)(7 - 3)(7 - 4)...*2*1

(7 - 2) appears twice, so it's not 7!, it's 7!*(7 - 2)

Maybe I missed something? Maybe it's supposed to be read in another way? I fully understand how we get from n! / [(n - r)!*r!) to [n(n - 1)(n - 2)...(n - [r - 1)] / r! , but this picture tells me how to get from the latter to the former, and I don't understand it!

Source:
http://gcecompilation.com/wp-content/uploads/2017/Pure%20Mathematics%20-1.pdf

page 136
Click to expand...

First, let's note that there is no real problem in the case you tried; you wrote an unnecessary factor of (7-2).

What it says, for n=7, r=3, is:

Code: 
[FONT=courier new][COLOR=#26282A]n(n - 1)...(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 =
[/COLOR][/FONT][FONT=courier new][COLOR=#26282A]7(7 - 1)...(7 - [3 - 1]) * (7 - 3)(7 - 3 - 1)*...*2*1 =
7(  6  )...(     5     ) * (  4  )(    3    )*...*2*1 = 7![/COLOR][/FONT]

The ellipses don't really do anything in this case.

A more typical example would be for n=9, r=4:

Code: 
[FONT=courier new][COLOR=#26282A]n(n - 1)...(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 =
[/COLOR][/FONT][FONT=courier new][COLOR=#26282A]9(9 - 1)...(9 - [4 - 1]) * (9 - 4)(9 - 4 - 1)*...*2*1 =
9(  8  )...(     6     ) * (  5  )(    4    )*...*2*1 = 9*8*7*6*5*4*3*2*1 = 9![/COLOR][/FONT]

Now, you are supposed to read such expressions intelligently, knowing that they are just an attempt to express a typical case, which might not make sense taken literally. For example, with n=5, r=2, it literally says this:

Code: 
[FONT=courier new][COLOR=#26282A]n(n - 1)...(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 =
[/COLOR][/FONT][FONT=courier new][COLOR=#26282A]5(5 - 1)...(5 - [2 - 1]) * (5 - 2)(5 - 2 - 1)*...*2*1 =
5(  4  )...(     4     ) * (  3  )(    2    )*...*2*1[/COLOR][/FONT]

where the 4 and the 2 are both duplicated. But it doesn't mean that; it means "start at 5 and continue down as far as you need to; at some point you will reach 4, and keep going; and end up with 1. Everything there except n, n-r, and 1 is only there to show that numbers are decreasing by 1 each time. This notation is just a way to avoid sigma notation, which is less natural to beginners. Don't let a notation that is meant to make things easier trip you up!

The important thing is that they are showing that if you multiply the numerator and denominator by the factorial of the next number down, (n-r)!, the resulting numerator is n!. Do you see that? For your example, this is

Code: 
[FONT=courier new]7*6   5*4*3*2*1    7*6*5*4*3*2*1     7!
--- * --------- = --------------- = ----
2*1   5*4*3*2*1   [FONT=courier new]2*1 * 5*4*3*2*1   2!5![/FONT][/FONT]
 
burgerandcheese said:
View attachment 10817

Hi

From the picture above, regarding the numerator only, how did they get n(n - 1)....(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 = n!
?

It just doesn't make sense to me. I tried it with real numbers too. Let n = 7, r = 3

7(7 - 1)(7- 1 - 1)(7 - [3 - 1])(7 - 3)(7 - 3 - 1])....*2*1
= 7(7 - 1)(7 - 2)(7 - 2)(7 - 3)(7 - 4)...*2*1

(7 - 2) appears twice, so it's not 7!, it's 7!*(7 - 2)

Maybe I missed something? Maybe it's supposed to be read in another way? I fully understand how we get from n! / [(n - r)!*r!) to [n(n - 1)(n - 2)...(n - [r - 1)] / r! , but this picture tells me how to get from the latter to the former, and I don't understand it!

Source:
http://gcecompilation.com/wp-content/uploads/2017/Pure%20Mathematics%20-1.pdf

page 136
Click to expand...
Note that the number after (n-r-1) is (n-r-1) + 1 = n-r. The number after n-r is n-r+1. The number after n-1 is n-1 + 1 or n
In reverse order n(n - 1)....(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1=1*2*...(n-r-1)(n-r)(n-(r-1))...(n-1)n = 1*2*...(n-r-1)(n-r)(n-r+1)...(n-1)n
The right hand side says to multiply 1 and and ALL the numbers up to and including n-r-1 and the the next number, n-r, and then the next number n-r+1 and all the numbers up to and including n-1 and finally by n. So which numbers between 1 and n was missed or duplicated??

n=7 and r=3: n(n - 1)....(n - [r - 1]) * (n - r)(n - r - 1)*...*2*1 = 7(7 - 1)....(7 - [3- 1]) * (7 - 3)(7 - 3 - 1)*...*2*1 = 7*6*5*4*3*2*1 = 7! No problem
 
OMG! I get it now :D


JeffM said:
\(\displaystyle \dbinom{7}{2} = \dfrac{7!}{2! * (7 - 2)!} = \dfrac{7 * 6 * 5!}{2! * 5!} = \dfrac{7 * 6}{2}.\)

More generally, \(\displaystyle \text {if } 0 \le r < n,\)

\(\displaystyle \dbinom{n}{r} = \dfrac{n!}{r! * (n - r)!} = \dfrac{\displaystyle \prod_{j=2}^n j}{\displaystyle r! * \left ( \prod_{j=2}^{n-r} j \right)} = \)

\(\displaystyle \dfrac{\displaystyle \left ( \prod_{j=2}^{n-r} j \right) * \left ( \prod_{j=(n+r)+1}^n j \right )} {\displaystyle \left ( \prod_{j=2}^{n-r} j \right) * r!} = \dfrac{\displaystyle \prod_{j=(n+r)+1}^n j}{r!}.\)
Click to expand...

Thank you for introducing me to this new symbol! :) in that last line, should it be from j = (n - r) + 1 instead of j = (n + r) + 1?
 
burgerandcheese said:
OMG! I get it now :D




Thank you for introducing me to this new symbol! :) in that last line, should it be from j = (n - r) + 1 instead of j = (n + r) + 1?
Click to expand...
Thank YOU for pointing out my error, which has now been fixed.

I normally am disgruntled when I make a careless error like that, but I am very not disgruntled or (quite gruntled according to Wodehouse) that you caught it because it does indeed mean that you understand the notation.

\(\displaystyle \Sigma\), which is the Greek capital letter sigma and stands for our letter S, stands for the first letter in Sum.

\(\displaystyle \Pi\), which is the Greek capital letter pi and stands for our letter P, stands for the first letter in Product.

Except for the difference in operation indicated, they work identically. They are succinct and clear (once you understand them).
 
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