A committee of 4people is to be chosen from 5 married couples. Find in how many ways the committee can be chosen if; 1) everyone is equally eligible. 2) the committee should include at least one woman
Without any restrictions, number of committees
= C(10,4) = 210
include at least one woman ---> exclude cases of all 4 being men
number of ways = 210 - C(5,4)
= 205
1) Ah, the classic committee conundrum! For this case, since everyone is equally eligible, we can simply select any combination of 4 people from the total of 10 individuals. Using a little math magic, we get the answer: C(10, 4) = 210. So, there are 210 ways to form the committee if everyone is equally eligible.
2) Now, time to spice things up by including at least one woman in the mix! To tackle this, we'll do a little subtraction trickery. First, let's calculate the total number of committees without restrictions (as we did in 1): C(10, 4) = 210.
Next, let's figure out the number of committees with all men: C(5, 4) = 5.
Now, by subtracting the number of all-men committees from the total, we get: 210 - 5 = 205. Voila! There are 205 ways to form the committee with at least one woman.
1) If everyone is equally eligible, we can select any 4 individuals from the total of 10 people (5 married couples).
So, the number of ways to choose the committee would be 10 choose 4.
Number of ways = C(10, 4) = 10! / (4!(10-4)!) = (10*9*8*7) / (4*3*2*1) = 210 ways.
Therefore, there are 210 ways to choose a committee of 4 people if everyone is equally eligible.
2) To find the number of ways to choose a committee that includes at least one woman, we need to consider two cases:
a) When the committee includes only one woman.
b) When the committee includes more than one woman.
a) When the committee includes only one woman:
We can choose 1 woman out of the 5 available and choose 3 people (both men and women) from the remaining 9 individuals (excluding the chosen woman).
Number of ways = C(5,1) * C(9,3) = 5 * (9! / (3! * (9-3)!)) = 5 * (9*8*7) / (3*2*1) = 5 * 84 = 420
b) When the committee includes more than one woman:
We can choose 2 women out of the 5 available and choose 2 people (both men and women) from the remaining 9 individuals.
Number of ways = C(5,2) * C(9,2) = (5! / (2! * (5-2)!)) * (9! / (2! * (9-2)!)) = 10 * 36 = 360
Therefore, the total number of ways to choose a committee that includes at least one woman = 420 + 360 = 780.
Therefore, there are 780 ways to choose a committee of 4 people that includes at least one woman.
To find the number of ways the committee can be chosen, we need to consider the conditions given for each case.
1) If everyone is equally eligible:
In this case, we can choose any 4 people from the 10 individuals (5 couples) without any restrictions. The order of selection does not matter for choosing the committee members.
To calculate the number of ways, we can use the combination formula: nCr = n! / (r!(n-r)!), where n is the total number of individuals (10) and r is the number of people to be chosen (4).
So, the number of ways to choose the committee when everyone is equally eligible is:
10C4 = 10! / (4!(10-4)!) = 10! / (4!6!) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1) = 210.
Therefore, there are 210 ways to choose the committee when everyone is equally eligible.
2) If the committee should include at least one woman:
We can consider two scenarios for this case: when one woman is selected and when more than one woman is selected. We can then calculate the total number of ways under these scenarios and add them together.
Scenario 1: Selecting one woman:
There are 5 options to choose one woman from the 5 couples. Once we have chosen one woman, we need to select the remaining 3 people from the remaining 9 individuals (excluding the chosen woman) without any restrictions.
Using the combination formula, the number of ways to choose the committee when one woman is selected is:
5C1 × 9C3 = 5! / (1!(5-1)!) × 9! / (3!(9-3)!) = 5 × (9 × 8 × 7) / (3 × 2 × 1) = 840.
Scenario 2: Selecting more than one woman:
We can select 2 women, 3 women, or all 4 women.
For selecting 2 women:
There are 5 options to choose the first woman and 4 options to choose the second woman from the remaining 4 couples (after the first woman is chosen). Once the two women are selected, we need to select the remaining 2 people from the remaining 8 individuals (excluding the two chosen women) without any restrictions.
Using the combination formula, the number of ways to choose the committee when 2 women are selected is:
5C2 × 8C2 = 5! / (2!(5-2)!) × 8! / (2!(8-2)!) = 10 × (8 × 7 × 6) / (2 × 1) = 840.
For selecting 3 women:
There are 5 options to choose the first woman, 4 options to choose the second woman, and 3 options to choose the third woman from the remaining 3 couples (after the first and second women are chosen). Once the three women are selected, we need to select the remaining 1 person from the remaining 7 individuals (excluding the three chosen women) without any restrictions.
Using the combination formula, the number of ways to choose the committee when 3 women are selected is:
5C3 × 7C1 = 5! / (3!(5-3)!) × 7! / (1!(7-1)!) = 10 × 7 = 70.
For selecting all 4 women:
There is only 1 option to choose all 4 women together. Once the four women are selected, we need to select the remaining 0 people since there are no more spots left in the committee.
Therefore, the number of ways to choose the committee when more than one woman is selected is:
840 + 840 + 70 + 1 = 1751.
Therefore, there are 1751 ways to choose the committee when the committee includes at least one woman.