Math Is Fun Forum

See my post #32.

Copy the url by right-clicking (or whatever you do on a 'smart'phone) on your Imgur image, and select 'Copy image address' (ie, the 'url'). When I did that on my PC, the copied url was 'https://i.imgur.com/ZVsL4f8.jpg'.

Then paste the url between 'img' (image) tags, to look like this:

[img]https://i.imgur.com/ZVsL4f8.jpg[/img]

Enter the result into your post, and the image should display. It does for me.

If the image display is too large (original size is 4608x3456 px, which is HUGE), refer to my post #18.

I get the following display sizes with your 4608x3456 px image:
(a) when not in a 'hide' box: across the width of the forum screen.
(b) when in a 'hide' box: nearly 4 times the width of the computer screen.

If you want to hide your image in a 'hide' box, enter the url as follows (but without the commas in the 'hide' brackets):

[hide,][img]https://i.imgur.com/ZVsL4f8.jpg[/img][/hide,]

Note: I entered the commas to render the 'hide' action inactive, otherwise a 'hide' box (containing the 'img' code) would have appeared instead of the 'hide' tags.

The 'img' tags are missing from your quotes of my post #18, which explains why the images don't show when I click on the links in your post.

So, it looks like 'img' tags vanish when quoting another post, which I'm pretty sure wasn't the case before the recent server change. And they vanish whether they're in a 'hide' box or not...but the 'hide' tags remain.

I don't have a smartphone and don't know much about them, but anyway, here's something that may help re this image problem...

I was able to view the image on my PC via the link from post #7. However, I suspect that the link isn't to the image itself, because Imgur image urls begin with 'https://i.imgur.com/' (different from nyc's 'http://imgur.com/gallery/'), and include a file extension (eg, '.jpg') after the filename (which nyc's doesn't).

That could account for the image not displaying here.

When I right-click the image in Imgur and select 'Copy image address', like so...

Entering that url between img tags successfully displays the image:

However, that image size is way too large for forum viewing (well, on my PC it is), so I shrank it to 640x480 for comparison:

The shrunk size fills just over half the width of my screen, which is about right for me.

The measurements given in the question are distances, not 'points' (exact locations) as such, and the error is in their placement within the 'points' notation (ie, the x,y elements in the brackets).

Hi mathland,

I was messing around with the 'Image of BBCode page' and changed the url, not realising that it would disable the link in your last post.

So if you particularly want to see the image version of that page, you'll now have to click on the link in my post instead of yours.

Sorry 'bout that, but I can't undo what I've done!

This is off-topic with post #1 here - I initially didn't look much beyond the thread title - but anyway, I'll just quickly say that 666 happens to be a Smith number.

Wikipedia: "In number theory, a Smith number is a composite number for which, in a given number base, the sum of its digits is equal to the sum of the digits in its prime factorization in the given number base."

For the number 666 (base 10):
     (a) The sum of its digits is 6+6+6 = 18
     (b) The sum of the digits in its prime factorisation (2x3x3x37) is 2+3+3+3+7 = 18
     And (a) = (b) = 18.

Hi Mitch,

I've just edited my post #58 and added a screen image of my Excel worksheet to show how I used MIF's Combinations Calculator output in Excel.

So you won't necessarily need to look at the YABASIC option...but it is interesting, and a totally different approach!

Btw, the idea for the YABASIC method came from the way I go about doing this longhand.

I tried MIF's Combinations and Permutations Calculator for the first time (understanding Cs & Ps has never been on my 'to do' list), and got it to work on the 2 scenarios I tried it on...with some issues (see end of post).

But at least they confirmed the total numbers of combinations mentioned in earlier posts, and spot checks I did on actual combo compositions were accurate.

These are my settings and outputs for the Post #1 and Post #55 scenarios:

In the 'List Them' entry, the first element in each scenario is the group comprising the fixed minimum number of players that must appear in each combo.

Without this 'trick' of reducing the number of elements, I failed in both scenarios:
   Post #1: My 'has 3' rule was rejected (the limit is 2, as Mitch said in post #13).
   Post #55: "List too large" error (MIF's max combinations is 10,000,000, which is well under the 86,493,225 for this scenario).

I copied MIF's output for post #55 and pasted it into Excel via the Text Import Wizard. That gave a single row/multiple column layout that I converted to multiple rows/multiple columns via Paste/Transpose. Player counts were then obtained by formula, producing the following:

To help explain some things, here's a screen image of my Excel worksheet:

Hi Mitch,

I got 56 combos:

I proceeded with the exercise by assuming that maybe you meant 'WRs' instead of 'PKs' where you said "Made up of a combination of QBs, RBs, TEs & PKs"...because PKs appears nowhere else.

Happy to redo my list if I made a false assumption.

Btw, some WRs in your list are below their minimum of 2.

In Bob's list in post #18 and mine in post #19 (both with 121 combinations), I noticed an arithmetic sequence when comparing the number of combinations between adjacent groups of p values.

1. The number of combinations of m & n (where both n+2m+3p ≤ 40 and n+m+p = 20) differ by 2 for adjacent p, eg:
     (a) {20n,0m,0p} to {0n,20m,0p} = 21
     (b) {19n,0m,1p} to {1n,18m,1p} = 19
     (c) {11n,0m,9p} to {9n,2m,9p} = 3
     (d) {10n,0m,10p} = 1 (the last option)

2. From 1 above, we get:
     1(a)-1(b) = 21-19 = 2, and
     1(c)-1(d) = 3-1 = 2
     ie, adjacent p {0p,1p} and (9p,10p} differ by 2.

3. There are 11 p groups (0p to 10p) in the full range (0p=21, 1p=19, 2p=17, 3p=15, 4p=13, 5p=11, 6p=9, 7p=7, 8p=5, 9p=3, 10p=1), and all adjacent p differ by 2.

4. Arithmetic sequence formula z(2x+y(z-1))/2 finds the total number of combinations, where:
     x = the first term...which is 1 (see 1(d) above)
     y = the 'common difference' between terms...which is 2 (see 2 & 3 above)
     z = how many terms to add up...which is 11 (see 3 above)

5. z(2x+y(z-1))/2 = 11(2*1+2(11-1))/2 = 121 combinations.

6. x, y & z in 4 above vary according to the number of slots (OP's = 20) and the sum total (OP's is ≤40).

Hi all;

I got Bob's list with these two codes (haven't tried it in Excel yet, nor looked at other ways):

The BASIC code can be run online at jdoodle.com's YABASIC, here.

Both codes test each case in the order produced from the following (a)+(b)+(c) nested sequence:
(a) each value of 3p (ie, p = 0 to 13), plus
(b) each value of 2m (ie, m = 0 to 20) for every (a), plus
(c) each value of n (ie, n = 0 to 20) for every (b)

The details of each result where both n+2m+3p<41 and n+m+p=20 are printed in the order found.

Edit: Since tried it in Excel...same answer: 121 combinations.