thuyduong91vnu wrote:
Hi Vyshak,
Thanks for your response
Yes I see the approach is applicable for the case in question (
which requires (1) a minimum of 4 members for each committee, and (2) odd number of members for committee 5 and 6), but I still wonder could it be so for all other cases.
Let's say in other cases (
in which we do not have to meet 2 requirements above), if we adjust the number of members for each commmittee (each of the first 5 committees has 1 member, the 6th committee has 21 members OR each of the first 5 committees has 2 member, the 6th committee has 16 members) such that the total number remains the same (= 26 members), then the answer could not be just \(\frac{26}{3}\) = 8 \(\frac{2}{3}\), right?
My question is quite confusing, isn't it
I just try to imagine another case and then apply the solution to see whether it works for all cases.
Thanks for your help!
Hi
thuyduong91vnu,
You are still not getting my point.
What does the question stem ask? --> Minimum number of board members necessary in order to staff all six committees
Given: 1) Board members must sit on a minimum of one committee and a maximum of three.
2) each committee must have a minimum of 4 members
3) Committees 5 and 6 must have an odd number of members
Now how do you minimize the number of members? --> By using the same member in 3 different committees --> By doing this we are counting each member 3 times. So we have to divide the total by 3. If each member could represent 4 committees then dividing by 4 will provide the answer, provided the question stem asks to minimize the number of members.
Now your question: adjust the number of members for each commmittee (each of the first 5 committees has 1 member, the 6th committee has 21 members OR each of the first 5 committees has 2 member, the 6th committee has 16 members) --> You have changed the conditions now. Let's take the case where 5 committees have 2 members and 5th committee has 16 members.
Total number = 26
Let A, B, C, D, E and F be the different committees --> Committee A, B and C can be formed by using just 2 distinct members.
D, E can be formed by using just 2 members. The members in D and E can represent one more committee F. So F will 14 distinct members
Total distinct members = (2 * 6)/3 + 4 --> 12/3 + 14 = 18
In the earlier case, the minimum number of each committee was 4: So assume A, B, C, D were made up of 4 members each and E, F were made of 5 members each
A, B, C can be formed by just 4 members
D can be formed 4 members
E, F will be formed by reusing the 4 members in D and adding 1 new member
Total distinct members = (4 * 6)/3 + 1 --> 24/3 + 1 = 9
So it all depends on how many times a member can be reused. A logic cannot be generalized to work in all scenarios.
Hope I am clear now and I answered your question