Question from this video: https://www.youtube.com/watch?v=nU2NrXOCdwk
Actually the word there was SUCCESSES, I will just change it to SUCCESSFUL.
So let me partially do the solution because I don't have a problem doing exactly the entire thing. From my understanding, you have to create integer partitions and those are represented by the cases.
SUCCESSFUL
Letter Count
- S – 3
- U – 2
- C – 2
- E – 1
- F – 1
- L – 1
- All – 10
Integer Partition Cases
- Case 1: {1, 1, 1, 1, 1}
- Case 2: {2, 1, 1, 1}
- Case 3: {2, 2, 1}
- Case 4: {3, 1, 1}
- Case 5: {3, 2}
To give an explanation, the integer partitions can be interpreted as
- Case 1 has all letters different from each other
- Case 2 has 2 letters identical to each other while the rest are all different
Remember that each set has to be taken from a pool of more letters.
In all honesty, I only know how to do well probably Case 1 and 2 and even then I'm not sure.
- Case 1: {1, 1, 1, 1, 1} – 5!
- Case 2: {2, 1, 1, 1} – (5C2 * 3C1) * 3!
Case 1 is pretty straight forward.
For case 2, I have used 5C2 to take two positions. Because those position will be occupied by identical letters, the order does not really matter. But I had to multiply by 3C1 because there are three possible letters.
Yeah the tricky part is that you don't consider just letters with two repetitions but those with at least 2, so S is included.
I'm at a loss on how to do Case 3 to 5. I have an idea but still not sure about it.
If anyone can comment on my solution if anything is wrong or suggest a better way then they may feel free to do so. Thank you.
Best Answer
The integer partition method will work, but you aren't counting it correctly.
Case 1. We are selecting 5 different letters from 6 choices of letters. 5! is what you get when you select 5 letters from 5 letters in a permutation, but there are 6 letters. Hence this should be 6 permute 5. Note that the number of 'C's or 'L's doesn't matter because the resulting word is the same no matter what 'C' or what 'S' you choose. Number of words = 6P5.
Case 2. 3C1 is indeed how to choose the 2 in {2,1,1,1}. But you also have to choose the 1's in {1,1,1}. After choosing the first number, there will be 5 numbers left. Hence, this will be equal to 5C3. There is one more step, which is to permute the word. The word is of length 5 with a repeated letter. The number of ways to permute this is 5!/2!. You can find more about this here: https://brilliant.org/wiki/permutations-with-repetition/. Number of words = (5C3 * 3C1) * 5!/2!.
In case 3, to choose the {2,2} from {2,2,1}, this will be 3C2. After selecting 2 numbers, there are 4 numbers left. Hence to select the 1 from {2,2,1}, this will be 4C1. Then, to permute a 5-letter word with 2 repeated letters, this is equal to 5!/(2! x 2!). Number of words = 3C2 * 4C1 * 5!/(2!2!).
Extending this method to case 4 and 5, in case 4 we should have 5C2 * 5!/3!, in case 5 we should have 2C1 * 5!/(3!2!). The total should be 5900.