Math Is Fun Forum

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.