Permutation of the word mathematics

Question

How many permutations are there of the letters Mathematics?

(a)How many of them begin and end with letter A?

(b)How many of them does not have two vowels adjacent to one another?

For (a) I got 9!/4

For (b) I do not even know how to start...

Answer 1

For part b, arrange the consonants MTHMTCS in $\frac {7!}{2!2!}$ ways and then arrange the vowels AEAI, together with XXXX, meaning four blanks or no vowels, in $\frac{8!}{2!4!}$ ways, into the 8 gaps before between and after the consonants. Then multiply the results.

Answer 2

Your answer for (a) seems correct to me.

To get you started on (b), try counting the number of permutations that do have two or more vowels adjacent to one another. Then subtract this number from the total number of permutations of mathematics to get the number of permutations that do not have two vowels adjacent to one another.

Answer 3

Your answer to the first question is correct. Since the first and last spaces are filled by A's, there are nine spaces to fill. We must fill two of them with M's, which can be done in $\binom{9}{2}$ ways. We must fill two of the remaining seven spaces with T's, which can be done in $\binom{7}{2}$ ways. The remaining five spaces can be filled with the five distinct remaining letters in $5!$ ways. Hence, the number of permutations of MATHEMATICS in which the first and last spaces are filled with A's is $$\binom{9}{2}\binom{7}{2} \cdot 5! = \frac{9!}{2!7!} \cdot \frac{7!}{2!5!} \cdot 5! = \frac{9!}{2!2!}$$

For the second problem, first arrange the seven consonants MTHMTCS. We choose two spaces of the seven spaces for the two M's, two of the remaining five spaces for the two T's, then arrange the remaining three distinct consonants in $3!$ ways. The number of such arrangements is $$\binom{7}{2}\binom{5}{2} \cdot 3! = \frac{7!}{2!2!}$$

For a particular arrangement of the consonants, we now have eight gaps in which to place the four vowels, six between successive consonants and two at the ends of the row. We can fill two of the eight gaps with A's in $\binom{8}{2}$ ways. We can fill one of the remaining six gaps with the E in six ways, then fill one of the remaining five gaps with the I in five ways. Hence, the number of ways of arranging the letters of the word MATHEMATICS so that no two vowels are adjacent is $$\binom{7}{2}\binom{5}{2} \cdot 3! \cdot \binom{8}{2} \cdot 6 \cdot 5 = \frac{7!}{2!2!} \cdot \frac{8!}{6!2!} \cdot 6 \cdot 5 \cdot \frac{4!}{4!} = \frac{7!}{2!2!} \cdot \frac{8!}{2!4!}$$