How-many-ways-can-the-letters-in-the-word-MATHEMATICS-be-rearranged-such-that-the-word-formed-neither-starts-nor-ends-with-a-vowel-and-any-four-consecutive-letters-must-contain-at-least-a-vowel-

Question Number 110729 by mr W last updated on 30/Aug/20

How many ways can the letters in

the word MATHEMATICS

be rearranged such that the word

formed neither starts nor ends with a

vowel, and any four consecutive

letters must contain at least a vowel ?

Answered by mr W last updated on 30/Aug/20

(s e e a l s o Q 110498)

n_{1} n_{2} n_{3} n_{4} n_{5}

= v o w e l

n_{1 \dots 5} = n u m b e r o f c o n s o n a n t s

1 ⩽ n_{1}, n_{5} ⩽ 3

0 ⩽ n_{2}, n_{3}, n_{4} ⩽ 3

n_{1} + n_{2} + n_{3} + n_{4} + n_{5} = 7

i n g e n e r a t i n g f u n c t i o n

{(x + x^{2} + x^{3})}^{2} {(1 + x + x^{2} + x^{3})}^{3}

= x^{2} {(1 + x + x^{2})}^{2} {(1 + x + x^{2} + x^{3})}^{3}

= \frac{x^{2} {(1 - x^{3})}^{2} {(1 - x^{4})}^{3}}{{(1 - x)}^{5}}

= x^{2} {(1 - x^{3})}^{2} {(1 - x^{4})}^{3} \underset{k = 0}{\sum^{\infty}} C_{4}^{k + 4} x^{k}

t h e c o e f f i c i e n t o f x^{7} t e r m i s :

C_{4}^{5 + 4} - 2 \times C_{4}^{2 + 4} - 3 \times C_{4}^{1 + 4} = 81

\Rightarrow 81 \times \frac{7!}{2! 2!} \times \frac{4!}{2!} = 1 224 720 w a y s

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