Exercise 5. A player has four black cards and three red cards, all distinct. In how many ways can he order them so that two successive cards are not both red?
Solution to exercise 5 We have the 10 arrangements $N R N R N R N, R N N R N R N, R N R N N R N$, RNRNRNN, NNRNRNR, NRNNRNR, NRNRNNR, RNNNRNR, RNNRNNR, $R N R N N N R$.

For each of these arrangements, we can permute the four black cards (24 possibilities) and permute the three red cards (6 possibilities), which gives a total of $10 \times 24 \times 6=$ 1440.