# 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?

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You count the number of ways they can sit down with the women in three consecutive seats (the number of good arrangements), and then you divide that by the number of ways they can sit down at all (the number of possible arrangements). In fact, nearly all of elementary probability can be summarized by the fraction $$\text{Probability} = \frac{\text{Good ways}}{\text{Possible ways}}$$ – Arthur Aug 7 '13 at 7:16
You count the total number of arrangments and for each possibility of three adjacent seats, you count the number of possibilities where the women are on those three seats. Can you take it from here? – Magdiragdag Aug 7 '13 at 7:16

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 $\frac{3! * 5!}{7!} = \frac{6 * 5!}{7 * 6 * 5!} = \frac{1}{7}$.

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Nice; I like the way you treat the group of three women as a block. – Magdiragdag Aug 7 '13 at 7:18
I don't understand the block thing – suffix Aug 7 '13 at 7:22
Once you have arranged the three women, if you only want to count arrangements in which they are together, you don't need to consider the possibility that they are apart. Thus, the three women can be considered one section of seating, since where one goes, they all go. – qaphla Aug 7 '13 at 7:23

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).

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