The number of arrangements, of the letters of the word MANAMA in which two M's do not appear adjacent, is

There are 6 letters. M repeats 2 times, A repeats 3 times. We first arrange all letters except $2 \mathrm{M}^{\prime} \mathrm{s}$ in $\frac{4!}{3!}$ $=4$ ways These 4 letters create 5 gaps, where we can arrange $2 \mathrm{M}^{\prime} \mathrm{s}$ in $\frac{{ }^5 \mathrm{p}_2}{2!}=10$ ways $\therefore \quad$ Required number of arrangements $=4 \times 10=40$