A grouping problem !!

[TheAnuja55]

Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?

[GMATGuruNY]

Orange Computers is breaking up its conference attendees into groups. Each group must have exactly one person from Division A, two people from Division B, and three people from Division C. There are 20 people from Division A, 30 people from Division B, and 40 people from Division C at the conference. What is the smallest number of people who will not be able to be assigned to a group?

This is a MAX/MIN problem.
To MINIMIZE the number of people not assigned to a group, we must MAXIMIZE the number of people who ARE assigned.

Each group is composed of the following ratio:
A : B : C = 1:2:3.

Since each group must include 3 of the 40 people from division C, the maximum number of people from Division C who could be assigned is 13:
13*3 = 39.

Multiplying A:B:C = 1:2:3 by 13, we get:
A : B : C = 13(1:2:3) = 13:26:39.

Thus:
Of the 20 people from Division A, the maximum who could be assigned = 13, the minimum not assigned = 20-13 = 7.
Of the 30 people from Division B, the maximum who could be assigned = 26, the minimum not assigned = 30-26 = 4.
Of the 40 people from Division C, the maximum who could be assigned = 39, the minimum not assigned = 40-39 = 1.

Minimum not assigned from the 3 divisions = 7+4+1 = 12.