A7. A) 210 For the two G-blocks, there are $6+5+4+3+2+1=21$ (or

$\left(\begin{array}{l}7 \\ 2\end{array}\right)$) possibilities to be placed on the 7 positions (see figure). For each choice, there are $4+3+2+1=10$ (or $\left(\begin{array}{l}5 \\ 2\end{array}\right)$) possibilities for the two N-blocks; then the three E-blocks are fixed. In total, therefore, $21 \times 10=210$.

Alternative solution: If the letter blocks were different, there would be 7! possible words. However, the three E-blocks are not different, so we count each word 3! times. Similarly for the two N-blocks and the two G-blocks. In total, therefore, $\frac{7 !}{3 ! \times 2 ! \times 2 !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 ! \times 4}=7 \times 6 \times 5=210$ different words.

![](https://cdn.mathpix.com/cropped/2024_04_17_7fefb8cd27a15e1d29a5g-2.jpg?height=555&width=271&top_left_y=75&top_left_x=1575)
A7. Seven letter blocks $G|\mathrm{E}| \mathrm{E} \mid \mathrm{N}$ are shuffled. Then you get, for example, $E|E| N|G| G$ or $G|E| N|E| N$.

How many different "words" of length 7 can be formed in total? (Any order of the 7 letters counts as a word.)
A) 210
B) 420
C) 840
D) 1260
E) 5040