Seven people (3 women and 4 men) arrange them selves randomly in seven consecutive seats in a row... Seven people (3 women and 4 men) arrange them selves randomly in seven consecutive seats in a row, find the probability the women will be in three adjacent seats.  
How to do this problem?
 A: There are $7!$ total ways of arranging the people.
If all three women sit together, then there are $3!$ ways of arranging just the women.
Then, the block of women and the four men need to get arranged. There are $5!$ ways to do this, as the block of women can be treated as one object to arrange, giving a total of five.
Thus, there are $3! * 5!$ ways that the people can be arranged such that the three women are together.
Therefore, the probability that the three women are together is $\frac{3! * 5!}{7!} = \frac{6 * 5!}{7 * 6 * 5!} = \frac{1}{7}$.
A: AAA can only be arranged one way. ABC can be arranged 6 different ways. We want all the woman sat together, so we treat them one thing/block, like the AAA. This is what is meant by treating them as one thing/block. However, that block of 3 women are individuals, unlike the AAA, so we need to multiply by 6 at the end. This is because all the permutations involving the men can be arranged with woman A first followed by woman B and then C; we can then swap woman A with woman B, giving BAC. We can do this 6 different ways (ABC, ACB, BAC, BCA, CAB, CBA), so this is why the block you have been treating as one thing multiplies the 5! at the end by 3! (6).
