For a while, I've been trying to remember how to figure out possible number of shaft-listing combinations given the number of shafts, especially for eight shafts. The Internet says, "A combination is the choice of r things from a set of n things without replacement and where order does not matter," and the formula is:

I studied this in school, I think we used the last notation, but I can't remember how to calculate. But we know, at the lower end:

2 shafts:

1, 2;

1&2 at once is invalid in weaving;

2 options.

3 shafts:

1, 2, 3;

1&2, 2&3;

1&2&3 at once is invalid in weaving;

5 options.

4 shafts:

1, 2, 3, 4;

1&2, 1&3, 1&4, 2&3, 2&4, 3&4;

1&2&3, 2&3&4, 1&3&4, 1&2&4;

1&2&3&4 at once is invalid in weaving;

14 options.

Can someone please remind me how to do this? Thank you.

EDIT: Thanks to a mysterious Fiona, we have an answer.

Not sure which part you are stuck in so will try and answer all parts.

I will write (n, r) to mean choose r from n.

So n! =1*2*3*...*n ie multiply all integers from 1 up to n together so

4! = 1*2*3*4=24 Then (4,3) =4!/(3!*1! )=4

Then for 4 shafts the answer is (4,1)+(4,2)+(4,3) = 4+6+4=14

For 8 shafts you want to calculate

(8,1)+(8,2)+(8,3)+(8,4)+(8,5)+(8,6)+(8,7).

Will leave you to do the math because I am on phone and out of time!

Hope that helps.

I have to sit with a pencil and paper to see if I really got it. I'm not sure where (n, r) came from, but the separate calculations is looking a lot like what we did in school. I'll get back to you, or you get back to me, once we figure this out, OK?

EDIT: So here goes, in language I'll understand when I revisit this post in future. You might want a nice drink, pencil, and paper, maybe a calculator about now.

First, I was reminded combination could also be expressed as "C(n,r)", which is why I knew what Fiona meant, but now I'm not sure which way we learned in school as I am also familiar with "!". Never mind, let's do weavers' counting. Let's figure out how many ways we can lift shafts on an eight-shaft loom. Staying with the last, fourth, notation on the top:

Numerator, the number on top:

n is the number of shafts, 8.

! is what's called "factorial" in English, and it wants you to multiply all integers from 1 to n.

n! is 8!=1*2*3*4*5*6*7*8=40,320. Whoa!

Now on to the Denominator, on the bottom:

r is the number of shafts I wish to lift at one time. In weaving, we don't do "all" or "none", but on a eight shaft loom, we can lift one, two, three, four, five, six, or seven shafts at once. And here's where it gets tricky; this math for Combination can only give us the number for each of these options. So let's go with lifting one shaft. r=1.

r! is 1!=1

(n-r)! is (8-1)!=7!=1*2*3*4*5*6*7=5,040.

r!(n-r)! is 1!(8-1)!=1*(1*2*3*4*5*6*7)=5,040.

40,320/5,040=8. This means there are eight way to lift just one shaft on an eight-shaft loom. And you might rightly shout, "Well, I could have told you that yesterday when you first posted this damn post!" True. But stay with me.

Conveniently, the number of options to lift only one shaft on an eight-shaft loom, 8, is the same as the number of options to lift seven shafts at once, 8. So we got two of seven combinations out of the way. I'll just go calculate the rest without bothering you, and let you know the answers.

Option for number of shafts to be lifted at once on an eight-shaft loom:

1 shaft at once, (same number of options as 7 shafts at once) = 8 options

2 shafts, (same number of options as 6 shafts) = 28 options

3 shafts, (same number of options as 5 shafts) = 56 options

4 shafts = 70 options

5 shafts, see 3 shafts above = 56 options

6 shafts, see 2 shafts above = 28 options

7 shafts, see 1 shaft above = 8 options

Add it all up, 8+28+56+70+56+28+8=254

On an eight-shaft loom, there are 254 ways to lift shafts.

"Why the heck did we need to know this?" I hear you ask. Well, I find this sort of thing fascinating, and I just have to find out. Sorry if I wasted your time. 254, though. I am impressed.

## 4 comments:

Not sure which part you are stuck in so will try and answer all parts.

I will write (n, r) to mean choose r from n.

So n! =1*2*3*...*n ie multiply all integers from 1 up to n together so

4! = 1*2*3*4=24

Then (4,3) =4!/(3!*1! )=4

Then for 4 shafts the answer is (4,1)+(4,2)+(4,3) = 4+6+4=14

For 8 shafts you want to calculate

(8,1)+(8,2)+(8,3)+(8,4)+(8,5)+(8,6)+(8,7).

Will leave you to do the math because I am on phone and out of time!

Hope that helps.

The individual calculations of ! parts definitely rings a bell, but slightly confused where the (n, r) came from, although that, too, rings a bell. Thank you, though, because I think I can work it out from here. Thank you very much, again. And Thanks. And ta.

Calculator!

https://www.omnicalculator.com/statistics/combination

Object is total number of shafts and sample is how many to lift at once.

Handy, Ben. Ta much. Used it to update the post.

Post a Comment