Question 857308: The organizing group which consists of 1 chairman, 1 financial secretary and 8 working staff is formed. The chairman and the finance secretary are chosen from 6 teachers while the 8 working staffs are selected from 20 students.
how many different organizing groups can be formed?
If it is possible to have an organizing group with all the working staff being boy, what are the possible numbers of boys among the 20 students?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you are choosing 1 chairman and 1 financial secretary from 6 teachers
this would be a permutation formula because order or position is important.
the formula would be 6P2 which is equal to 6! / (6-2)! which is equal to 30.
you want to choose 8 staff members from 20 students.
this would be a combination formula because order or position is not important.
the formula would be 20C8 which is equal to 20! / (8! * (20-8)! which is equal to 125970.
the 30 different combinations of teachers and the 125970 combinations of working staffs combine to make a total of 30 * 125970 possible overall combinations which is equal to 3779100.
if you wanted the working staff to be all boys, you would need a minimum of 8 boys as members of the student body. I'm not sure I understand what you are looking for with this question. In that case you would be choosing from the 8 boys and you would only have one choice. If there were 9 boys as members of the student body, then you would be choosing 8 from 9 which would be the combination formula of 9C8. Similarly, if 10 boys as members of the student body, you would be choosing 8 from 10 which would be the combination formula of 10C8, etc.
I you want to see how the general questions worked in practice, you would need much smaller numbers so the individual results could be tabulated.
For example:
Assume 3 teachers for the positions of chairman and secretary.
These could be formed in 3*2 = 6 ways.
Assume the teachers are a,b,c,d,e,f.
The 6 ways of getting a chairman and a secretary are:
a,b
a,c
a,d
a,e
plus the reverse of:
b,a
c,a
d,a
e,a
this is because order or position is important.
a,b means a is the chairman and b is the secretary.
b,a means b is the chairman and a is the secretary.
So we are talking about 6 possible combinations for chairman and secretary where order is important.
now suppose you have 3 students and you want to choose 2 working staff from them.
These could be formed in 3*2/2 ways which is equal to 3 because order is not important.
Assume the students are numbers 1,2,3
The possible ways are:
1,2
1,3
2,3
The reverse of 2,1 and 3,1 and 3,2 are not counted because order is not important and 1,2 and 2,1, and 1,3 and 3,1, and 2,3 and 3,2 are the same students so you would not want to count them twice as you would if order or position was important.
So you have 6 possible ways to get a chairman and a secretary and 3 possible ways to get 2 working staff.
The total number of possible combinations is therefore 6 * 3 = 18 possible different overall combinations.
Those are:
a,b with 1,2
a,c with 1,2
a,d with 1,2
a,e with 1,2
a,b with 2,3
a,c with 2,3
a,d with 2,3
a,e with 2,3
a,b with 1,3
a,c with 1,3
a,d with 1,3
a,e with 1,3
plus the reverse of:
b,a with 1,2
c,a with 1,2
d,a with 1,2
e,a with 1,2
b,a with 2,3
c,a with 2,3
d,a with 2,3
e,a with 2,3
b,a with 1,3
c,a with 1,3
d,a with 1,3
e,a with 1,3
that's a total of 18 possible combinations which agrees with what the formula says it would be.
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