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Question Number 110498 by Aina Samuel Temidayo last updated on 29/Aug/20

How many ways can the letters in the word MATHEMATICS be rearranged such that the word formed either starts or ends with a vowel, and any three consecutive letters must contain a vowel?

HowmanywayscanthelettersinthewordMATHEMATICSberearrangedsuchthatthewordformedeitherstartsorendswithavowel,andanythreeconsecutivelettersmustcontainavowel?

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

I have the options. 332640 is not part of them.

Ihavetheoptions.332640isnotpartofthem.

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

It′s not also part of the options.

It′snotalsopartoftheoptions.

Commented by Aina Samuel Temidayo last updated on 29/Aug/20

That′s part of the options. So can you send the solution?

That′spartoftheoptions.Socanyousendthesolution?

Answered by mr W last updated on 29/Aug/20

we have four vowels: A,A,E,I seven consonants: M,M,T,T,H,C,S any three consecutive letters must contain at least a vowel, that means two vowels may separated by at most two consonants. a valid word should start with a vowel or end with a vowel. let′s look the first case: it starts with a vowel and ends with a consonant. such a valid word can be formed like: ■□■□■□■♦ ■ stands for one of the 4 vowels □ can be occupied by none, one or two consonants ♦ can be occupied by one or two consonants to occupy the three □ places and the one ♦ place with seven consonants, there are 4 ways, they are ■X■XX■XX■XX ■XX■X■XX■XX ■XX■XX■X■XX ■XX■XX■XX■X or using generating function: (1+x+x^2 )^3 (x+x^2 ) the coefficient of x^7 term is 4, that means there are 4 ways to occupy the places with the 7 consonants. to arrange the 4 vowels there are ((4!)/(2!)) ways. to arrange the seven consonants there are ((7!)/(2!2!)) ways. so we have 4×((4!)/(2!))×((7!)/(2!2!)) ways to form a word which starts with a vowel and ends with a consonant. we have the same number of ways to form a word which ends with a vowel and starts with a consonant. therefore totally we have 2×4×((4!)/(2!))×((7!)/(2!2!))=120960 ways to form a word as requested.

wehavefourvowels:A,A,E,Isevenconsonants:M,M,T,T,H,C,Sanythreeconsecutivelettersmustcontainatleastavowel,thatmeanstwovowelsmayseparatedbyatmosttwoconsonants.avalidwordshouldstartwithavowelorendwithavowel.let′slookthefirstcase:itstartswithavowelandendswithaconsonant.suchavalidwordcanbeformedlike:◼◻◼◻◼◻◼♢◼standsforoneofthe4vowels◻canbeoccupiedbynone,oneortwoconsonants♢canbeoccupiedbyoneortwoconsonantstooccupythethree◻placesandtheone♢placewithsevenconsonants,thereare4ways,theyare◼X◼XX◼XX◼XX◼XX◼X◼XX◼XX◼XX◼XX◼X◼XX◼XX◼XX◼XX◼Xorusinggeneratingfunction:(1+x+x2)3(x+x2)thecoefficientofx7termis4,thatmeansthereare4waystooccupytheplaceswiththe7consonants.toarrangethe4vowelsthereare4!2!ways.toarrangethesevenconsonantsthereare7!2!2!ways.sowehave4×4!2!×7!2!2!waystoformawordwhichstartswithavowelandendswithaconsonant.wehavethesamenumberofwaystoformawordwhichendswithavowelandstartswithaconsonant.thereforetotallywehave2×4×4!2!×7!2!2!=120960waystoformawordasrequested.

Commented by Aina Samuel Temidayo last updated on 30/Aug/20

Thanks.

Thanks.

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