A committee of 4 people is to be chosen from 5 married coiples.find in how many ways the committee can be chosen if:
i) everyone is equallt eligible
ii) the committee should include at least one woman
solution so far.5 couples=10 people.
To choose 4=10C4=10!/(10-4)!4!=210 ways.pls help confirm if i am correct and also help solve (ii)
You are correct in your solution for part (i). The number of ways to choose a committee of 4 people from 10 individuals (5 married couples) is indeed 210, which can be calculated using the combination formula.
Now, let's move on to part (ii) where the committee should include at least one woman. To solve this, we can break it down into two cases:
Case 1: The committee consists of exactly one woman.
In this case, we need to choose 1 woman out of 5 (the 5 married couples) and 3 other individuals from the remaining pool of 9 (since we've already selected one woman). This can be calculated as 5C1 * 9C3, which equals 5 * 84 = 420.
Case 2: The committee consists of more than one woman.
In this case, we choose 2, 3, or 4 women from the available 5 women, and the remaining committee members are selected from the remaining pool of 9 (since we've already selected some women).
- If we choose 2 women: 5C2 * 9C2 = 10 * 36 = 360.
- If we choose 3 women: 5C3 * 9C1 = 10 * 9 = 90.
- If we choose 4 women: 5C4 * 9C0 = 5 * 1 = 5.
To get the total number of ways, we add up the cases where we have exactly one woman, more than one woman, and no women. So, the total number of ways the committee can be chosen is 420 + 360 + 90 + 5 = 875.
Therefore, the committee can be chosen in 875 ways if it should include at least one woman.
i) 210 is correct
ii) So what we don't want is to have all-men committees
number of all men committees = C(5,4) = 5
so the number with at least one woman
= 210 - 5 = 205