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?

Search

9 May
2017

Bruno Zindyonok…i try for the love of math…420?

Rajib Royonshould be more

Bruno Zindyonoh yes sorry i did not read the question correctly…hmm that is tougher

Bruno Zindyon1680? else i will leave it to more clever guys…

Dhananjay NeneonIt’s so much easier to write a brute force program than recollecting permutation combination formulae

Rajib RoyonWhere were you when I was struggling in math in high school? ðŸ™‚

Manajit Senguptaon8!/3!2!2! = 1680

Dhananjay NeneonYep, thats correct. And although I didn’t know enough to solve it that way, here’s the brute force proof ðŸ™‚ ðŸ™‚

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

Anand Iyeron8!/2?

Rajib RoyonProbably far less because of the repeats

Anand IyeronIf you are always making 8 letter words, then half of the permutations have “m” before “e”

Rajib RoyonThat part is indeed correct

Manajit Senguptaon8!/3!2! – 7!

Rajib RoyonThat is probably too high

Manajit SenguptaonActually my first answer is negative – 1680 seems to be right 8!/3!2!2

Anand IyeronBut i also see that semantically two eight letter words with repeat letters in similar positions are in fact not unique..

Rajib RoyonHow do? MOON and MOON is the same although the Os are interchanged, right?

Anand IyeronBut MOON is a 4 letter word, bad requirements…

Anand IyeronThis is scope creep personified

Rajib RoyonYou would prefer I put a R in the middle? ðŸ™‚

Anand IyeronWorse new problem..that is litigation territory

Balaji Kaneon8!/(3!*2!) – 7!/(3!*2!)

Rajib RoyonI get the first part. How did you come up with the second?

Shiva ShenoyonRajib Roy Subtracting the case where the E is before the M maybe?

Rajib RoyonAlmost there.

Rajib RoyonAlmost there

Balaji KaneonThe second part is to account for all cases where EM appear together. Let me see what else I may be missing !

Sundar Ramon1680? Half the permutations would have m before e and vice versa.

Somnath Daripaon2940

Rajib RoyonToo high

Ranajoy GangulionJust wrote 21 words without caring for maths. There could be many more but I am sleepy now.

Sudipto Ghoseon28x6C3x3C2