Answer: 21600.

Solution. First, arrange the vowels, among which there are two letters E. They can be permuted in $5!/2=60$ ways. Now, 6 positions are formed where consonants can be placed (no more than one consonant per position) - these are the four gaps between the vowels, as well as the positions at the beginning and the end. The letter $\Gamma$ can be placed in any of the 6 positions. Then, there are 5 positions available for M. After that, we choose a position for $\mathrm{T}$ in 4 ways, and finally for $\mathrm{P}$ in 3 ways. By the rule of product, the total is $60 \cdot 6 \cdot 5 \cdot 4 \cdot 3=21600$.

Grading. 7 points for a correct solution. Minus one point for an arithmetic error (with correct reasoning). If the number of permutations of the vowels is calculated incorrectly, but the second part of the solution is correct, 2 points.
3. How many ways are there to rearrange the letters of the word ГЕОМЕТРИЯ so that no two consonants stand next to each other?

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