9 May
2017
Here is a puzzle that will remind you of math from high school
Take the word “MONOTONE”. In how many different ways can you jumble up all the letters such that M will never come after E?
Take the word “MONOTONE”. In how many different ways can you jumble up all the letters such that M will never come after E?
ok…i try for the love of math…420?
should be more
oh yes sorry i did not read the question correctly…hmm that is tougher
1680? else i will leave it to more clever guys…
It’s so much easier to write a brute force program than recollecting permutation combination formulae
Where were you when I was struggling in math in high school? 🙂
8!/3!2!2! = 1680
Yep, thats correct. And although I didn’t know enough to solve it that way, here’s the brute force proof 🙂 🙂
print(len(filter(lambda s: s.index(‘M’) < s.index('E'),
set(itertools.permutations(
['M','O','N','O','T','O','N','E'])))))
Its in a language called python
8!/2?
Probably far less because of the repeats
If you are always making 8 letter words, then half of the permutations have “m” before “e”
That part is indeed correct
8!/3!2! – 7!
That is probably too high
Actually my first answer is negative – 1680 seems to be right 8!/3!2!2
But i also see that semantically two eight letter words with repeat letters in similar positions are in fact not unique..
How do? MOON and MOON is the same although the Os are interchanged, right?
But MOON is a 4 letter word, bad requirements…
This is scope creep personified
You would prefer I put a R in the middle? 🙂
Worse new problem..that is litigation territory
8!/(3!*2!) – 7!/(3!*2!)
I get the first part. How did you come up with the second?
Rajib Roy Subtracting the case where the E is before the M maybe?
Almost there.
Almost there
The second part is to account for all cases where EM appear together. Let me see what else I may be missing !
1680? Half the permutations would have m before e and vice versa.
2940
Too high
Just wrote 21 words without caring for maths. There could be many more but I am sleepy now.
28x6C3x3C2
Number of ways in which M always comes before E = 7.(7+1)/2 = 28
Arranging the rest of the six letter for each of the above cases = 6!/(3!.2!) = 60
Therefore, the answer is = 28.60 = 1680
That works! Another way to think about it… combinations of 8 objects with 3 like and 2 like objects – 8! / (3! x 2! )
Half of them will have M before E and vice versa