Out of 4 architects, 6 engineers and 9 electricians, how many ways can a group of 6 people be formed to attend a forum if:
a. there is an equal number of architects, engineers and electricians
b. two particular engineers cannot be in the committee?
c. there are at least 3 electricians?
a. equal number of each
Combinations again..
architects 4!/(2!2!)
engineers 6!/(2!4!)
electricians 9!/(2!7!)
b. remove the 2 engineers from the choices which would leave 4 architechts, 4 engineers and 9 electricians Now, we have 17 to choose from to make a group of 6.
17!/(6! 11!)
Thanks much :)
To find the number of ways a group of 6 people can be formed to attend a forum, we need to apply the concept of combinations.
a. When there is an equal number of architects, engineers, and electricians:
Since we need to select an equal number of each profession, we will select 2 architects, 2 engineers, and 2 electricians.
The number of ways to select 2 architects out of the available 4 is represented as C(4, 2) or 4C2, which is calculated as (4!)/(2!(4-2)!) = 6.
Similarly, the number of ways to select 2 engineers out of the available 6 is C(6, 2) or 6C2, calculated as (6!)/(2!(6-2)!) = 15.
And the number of ways to select 2 electricians out of the available 9 is C(9, 2) or 9C2, calculated as (9!)/(2!(9-2)!) = 36.
To find the total number of ways, we multiply these three results together: 6 * 15 * 36 = 3,240.
Therefore, when there is an equal number of architects, engineers, and electricians, there are 3,240 ways to form a group of 6 people to attend the forum.
b. When two particular engineers cannot be in the committee:
There are two engineers who cannot be selected, meaning we need to subtract these cases from the total number of ways.
First, we calculate the total number of ways without any restrictions. This is the same as part (a), which gives us 3,240 ways.
Now, we calculate the number of committees that include the two particular engineers. Since both engineers need to be in the committee, we select the remaining 4 people from the available candidates: 4 architects, 4 engineers (excluding the two particular ones), and 9 electricians.
The number of ways to select 4 people out of the 4 architects is C(4, 4) or 4C4, which is calculated as (4!)/(4!(4-4)!) = 1.
Similarly, the number of ways to select 4 people out of the 4 engineers (excluding the two particular ones) is C(4, 4) or 4C4, calculated as (4!)/(4!(4-4)!) = 1.
And the number of ways to select 4 people out of the 9 electricians is C(9, 4) or 9C4, calculated as (9!)/(4!(9-4)!) = 126.
To find the total number of ways with the restriction, we multiply these three results together: 1 * 1 * 126 = 126.
Finally, we subtract the number of ways with the restriction from the total number of ways: 3,240 - 126 = 3,114.
Therefore, when two particular engineers cannot be in the committee, there are 3,114 ways to form a group of 6 people to attend the forum.
c. When there are at least 3 electricians:
To calculate this, we will consider different cases:
- Selecting exactly 3 electricians and 3 from the remaining candidates.
- Selecting exactly 4 electricians and 2 from the remaining candidates.
- Selecting exactly 5 electricians and 1 from the remaining candidates.
- Selecting all 6 electricians.
Case 1: Selecting exactly 3 electricians and 3 from the remaining candidates:
The number of ways to select 3 electricians out of the 9 available is C(9, 3) or 9C3, calculated as (9!)/(3!(9-3)!) = 84.
Now, we need to select 3 more people from the remaining candidates: 4 architects, 6 engineers, and the remaining 6 electricians (excluding the 3 already selected).
The number of ways to select 3 people out of the 4 architects is C(4, 3) or 4C3, calculated as (4!)/(3!(4-3)!) = 4.
Similarly, the number of ways to select 3 people out of the 6 engineers is C(6, 3) or 6C3, calculated as (6!)/(3!(6-3)!) = 20.
To find the total number of ways for this case, we multiply these results together: 84 * 4 * 20 = 6,720.
Case 2: Selecting exactly 4 electricians and 2 from the remaining candidates:
The number of ways to select 4 electricians out of the 9 available is C(9, 4) or 9C4, calculated as (9!)/(4!(9-4)!) = 126.
Now, we need to select 2 more people from the remaining candidates: 4 architects, 6 engineers, and the remaining 5 electricians (excluding the 4 already selected).
The number of ways to select 2 people out of the 4 architects is C(4, 2) or 4C2, calculated as (4!)/(2!(4-2)!) = 6.
Similarly, the number of ways to select 2 people out of the 6 engineers is C(6, 2) or 6C2, calculated as (6!)/(2!(6-2)!) = 15.
To find the total number of ways for this case, we multiply these results together: 126 * 6 * 15 = 11,340.
Case 3: Selecting exactly 5 electricians and 1 from the remaining candidates:
The number of ways to select 5 electricians out of the 9 available is C(9, 5) or 9C5, calculated as (9!)/(5!(9-5)!) = 126.
Now, we need to select 1 more person from the remaining candidates: 4 architects, 6 engineers, and the remaining 4 electricians (excluding the 5 already selected).
The number of ways to select 1 person out of the 4 architects is C(4, 1) or 4C1, calculated as (4!)/(1!(4-1)!) = 4.
Similarly, the number of ways to select 1 person out of the 6 engineers is C(6, 1) or 6C1, calculated as (6!)/(1!(6-1)!) = 6.
To find the total number of ways for this case, we multiply these results together: 126 * 4 * 6 = 3,024.
Case 4: Selecting all 6 electricians:
The number of ways to select all 6 electricians out of the 9 available is C(9, 6) or 9C6, calculated as (9!)/(6!(9-6)!) = 84.
To find the total number of ways for this case, we multiply this result by 1 (as we don't need to select anyone else): 84 * 1 = 84.
Finally, we add up the results from all four cases to get the total number of ways: 6,720 + 11,340 + 3,024 + 84 = 21,168.
Therefore, when there are at least 3 electricians, there are 21,168 ways to form a group of 6 people to attend the forum.