6 people are sitting around a table. Let x be the number of people sitting next to at least one woman and y be the number of people sitting next to at least one man. How many possible values of the ordered pair (x,y) are there? (For example, (6,0) is the pair if all 6 people are women, since all 6 people are sitting next to a woman, and 0 people are sitting next to a man.)
Really hard. Can I have some help?
The number of ordered pairs (x,y) is 12.
Of course! Let's break down the problem and see how we can approach it.
First, let's consider the number of possible values for x. In order for someone to be sitting next to at least one woman, there must be at least one woman next to them. We can start by examining the possible configurations for the women.
Case 1: Only one woman is seated.
In this case, we can place the woman in any of the 6 seats. Since the remaining 5 people can be anyone (men or women), there are 6 possible configurations.
Case 2: Two women are seated.
We can choose the positions for the two women in C(6, 2) = 15 ways, where C(n, r) represents the number of combinations of selecting r items from a set of n items without replacement. Now, since the remaining 4 people can be anyone, there are 2^4 = 16 possible configurations for them.
Case 3: Three women are seated.
Similarly, we can choose the positions for the three women in C(6, 3) = 20 ways. With the remaining 3 people, there would be 2^3 = 8 possible configurations.
Note that if there are more than three women seated, the number of possible configurations for x will be the same as the case of three women seated (since we are only interested in whether someone is seated next to at least one woman or not).
Now, let's consider the number of possible values for y. The same logic applies here – we can analyze different cases for the number of men seated.
Case 1: Only one man is seated.
As in the case of one woman, there are 6 possible configurations for the man, and the remaining 5 people can be anyone, resulting in 6 possible configurations.
Case 2: Two men are seated.
We can choose the positions for the two men in C(6, 2) = 15 ways, and the remaining 4 people can be anyone, resulting in 2^4 = 16 possible configurations.
Case 3: Three men are seated.
We can choose the positions for the three men in C(6, 3) = 20 ways, and the remaining 3 people can be anyone, resulting in 2^3 = 8 possible configurations.
Again, if there are more than three men seated, the number of possible configurations for y will be the same as the case of three men seated.
To find the total number of possible values for the ordered pair (x, y), we need to consider all combinations of x and y from the different cases we discussed.
From the cases we calculated, we have:
- 1 configuration for (x, y) = (1, 1) (one woman seated next to one man)
- 6 configurations for (x, y) = (1, y), where y is between 0 and 6 (one woman seated next to any number of men)
- 6 configurations for (x, y) = (x, 1), where x is between 0 and 6 (one man seated next to any number of women)
- 90 configurations for (x, y) = (x, y), where both x and y are between 0 and 6 (any combination of women and men seated)
Adding these up, we find that there are 1 + 6 + 6 + 90 = 103 possible values for the ordered pair (x, y).