A committee of 6 people is to be selected randomly from a group of 11 men and 12 women. Find the number of possible committees if:
a) a particular woman is included
b) a particular man is not included
c) there are to be 4 women and 2 men
a) Hmmm, including a particular woman, eh? Well, there are 11 men and 12 women in the group. So, if a particular woman is included, we can choose the remaining 5 people from the rest of the group, which consists of 10 men and 11 women. That gives us a total of (10+11)C5 possible committees. That's (21C5).
b) "A particular man is not included." Hmm, sounds like someone might have rubbed this particular man the wrong way. So, if this grumpy man is not included, we can choose all 6 committee members from the remaining 10 men and 12 women. That gives us (10+12)C6 possible committees. That's (22C6).
c) Alright, if we want to have 4 women and 2 men on the committee, we need to choose those 4 women from 12, which gives us (12C4) possibilities. Similarly, we need to choose the 2 men from 11, which gives us (11C2) possibilities. Multiply them together, and voila! That'll give you the total number of possible committees.
Remember, my friend, to use those binomial coefficient formulas to calculate the combinations. Happy counting!
a) To find the number of possible committees if a particular woman is included, we need to select 5 more people from the remaining 11 men and 11 women.
The number of ways to select 5 people from a group of 11 men and 11 women can be calculated using combinations. We can use the formula nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items being selected.
In this case, we have 11 men and 11 women, so the number of ways to select 5 people is:
11C5 = 11! / (5!(11-5)!) = (11 * 10 * 9 * 8 * 7) / (5 * 4 * 3 * 2 * 1) = 462
Therefore, there are 462 possible committees if a particular woman is included.
b) To find the number of possible committees if a particular man is not included, we need to select 6 people from the remaining 10 men and 12 women.
The number of ways to select 6 people from a group of 10 men and 12 women can be calculated using combinations:
(10 + 12)C6 = 22C6 = 22! / (6!(22-6)!) = (22 * 21 * 20 * 19 * 18 * 17) / (6 * 5 * 4 * 3 * 2 * 1) = 746,13
Therefore, there are 746,13 possible committees if a particular man is not included.
c) To find the number of possible committees with 4 women and 2 men, we need to select 4 women from the group of 12 and 2 men from the group of 11.
The number of ways to select 4 women from a group of 12 can be calculated using combinations: 12C4 = 12! / (4!(12-4)!) = (12 * 11 * 10 * 9) / (4 * 3 * 2 * 1) = 495
The number of ways to select 2 men from a group of 11 can be calculated using combinations: 11C2 = 11! / (2!(11-2)!) = (11 * 10) / (2 * 1) = 55
To find the total number of possible committees with 4 women and 2 men, we multiply the number of ways to select the women by the number of ways to select the men:
495 * 55 = 27,225
Therefore, there are 27,225 possible committees with 4 women and 2 men.
To solve these problems, we need to use the concept of combinations.
The number of combinations of selecting r items from a set of n items is given by the formula:
nCr = n! / (r!(n-r)!)
Where "n" is the total number of items and "r" is the number of items to be selected.
a) If a particular woman is included:
Since we need to select a particular woman, we have fixed one position in the committee for her. Therefore, we only need to select the remaining 5 members from the remaining 11 men and 11 women.
The number of possible committees in this case = 5C5 * 11C5
b) If a particular man is not included:
Similarly, we have fixed one position for the particular man who is not included. We need to select the remaining 5 members from the remaining 11 men and 12 women.
The number of possible committees in this case = 5C6 * 12C5
c) If there are to be 4 women and 2 men:
We need to select 4 women from 12 women and 2 men from 11 men.
The number of possible committees in this case = 4C4 * 12C4 * 2C2 * 11C2
To find the values of these combinations, you can use mathematical calculation or a calculator with factorial function to calculate the factorials and then substitute them in the formulas. Alternatively, you can use online combination calculators or specific software that can calculate combinations for you.
a) Once the particular woman is in, you need 5 more of the remaining 22 people
---> C(11,5) =
b) one man is out, so you only have 22 people, choose 6 of them
---> C(22,6) = ..
c) ----> C(12,4) x C(11,2) = ....