Math Is Fun Forum

If I understood correctly, your 228 combinations contain 18 combinations with double "O"?

Still, if they share only sides, a corner-point has 2 adjacent points, an edge-point has 3, and an inside one has 4. Maybe, in the end, I don't understand the meaning of the word "adjacent", being a non-native English speaker...

Never mind, I didn't mean to interfere, just wanted to help him out a bit. Sorry if I caused any confusion.

Okay, point taken.
I still prefer reasoning though, whenever possible.

I assume many members are just interested in getting the correct answers followed by a simple explanations they can actually understand. On the other side, advanced users are probably interested in learning more, and would benefit from the gf.

Just saying that not all of us on here are advanced users, or interested in the gf. It would be then reasonable to think about adding a small feature, like a question to be answered by new members for instance, so that one can say what exactly we're looking for when posting a question in "Help me !" section.

I don't really understand why people use all that in post #2 for combinatorics problems as simple as this one.
And I don't see how come it's mandatory, unless I'm missing something...

There is A LOT easier way to solve this, without using any formulas at all.
Maybe someone will find it useful, so I'm going to explain in plain English

The word has 8 letters. If they all were different letters, our solution would be 8*7*6=336 (first choose one letter out of eight, then one out of the remaining seven, and one out of the remaining six letters).

But! We want only words that contain three DIFFERENT letters.

Letter "O" appears twice, so we need first to eliminate all the combinations that contain a double "O".

This double "O" can be combined with each of the 6 remaining letters; additionally, the double "O" can have three different positions in a three-letter word (i.e. OO-, O-O, -OO). So that's 6*3 combinations that contain the double "O" and that we need to subtract from 336.

Then, we also need to account for 1 "O" overcount in our original result. This means that in some 3-word combinations the first O (O1) was used and in some the second O (O2) was used, but these are essentially the same words (MO1N and MO2N = MON). So, we need to eliminate this as well.

We choose either O1 or O2 to eliminate. Again, this one "O" is combined with 2 other non-O letters (first, we choose one letter out of 6 non-O letters, then one out of the remaining 5 letters); additionally, "O" can have three different positions in a word (i.e. O--, -O-, --O). So that's 6*5 combinations for one position multiplied by 3 positions: 30*3=90, meaning we need to subtract 90 additional combinations.

Finally:

336-18-90 = 228

Hope someone will find this helpful.

OK, I'm pretty sure this can work like this, using my patterns, just need to check once again I've got all the patterns, then it's relatively easy to find the combinations.

Thanks again for your help.

If someone else finds a neat way to solve this in the meantime, please do let me know.

Here's what I've come up with:

-Firstly, based on these 15 sets of four, there is a distribution pattern:
item 1 - position 1 (10)
item 2 - position 1 or 2 (4+6)
item 3 - position 1 or 2 or 3 (1+6+3)
item 4 - position 2 or 3 or 4 (3+6+1)
item 5 - position 3 or 4 (6+4)
item 6 - position 4 (10)

So, items 1 and 6 never change their position in any set of four.

- Secondly, possible positions for items 2-5 can be:
24 / 42
33
123 / 132 / 231 / 213 / 312 / 321

Thirdly, taking into account the fixed positions in the first step, all possible positions are:
item 1 - 1
item 2 - 24 / 42 / 33
item 3 - 123 / 132 / -42 / -33
item 4 - 231 / 321 / 24- / 33-
item 5 - 24 / 42 / 33
item 6 - 4

* e.g. distribution of items 4 (6 in total, in 9 sets of four), based on its position in each set (a set has four possible places), can be, say, "24-"  meaning 2 items in the second place, 4 items in the third place, none in the fourth place (and none in the first place)

Now, please tell me if this makes sense...
I forgot how to use formulas, high school was a long time ago...

Hi bobbym,

Thanks for your prompt reply!

There was a typo in my previous post, I need to choose 9 sets out of 15 (instead of 6; obviously, it is not possible to choose 6 sets (of four items) which would contain each item six times in the end).

So, the sets (of 4) are fixed after the first step.
These are ("1" representing the item 1, "2" representing the item 2, etc.):

1234
1235
1236
1245
1246
1256
1345
1346
1356
1456
2345
2346
2356
2456
3456

Now, I need to see how many ways I can choose 9 sets to have each item six times in the end.
For instance, this is one solution:

1234
1235
1246
1256
1345
1456
2346
2356
3456

Hope it's a bit clearer.

Thank you very much for your help!