4 different speakers:GmatKiss wrote:An oratorical society consists of six members, and at an upcoming meeting, the members will present a total of four speeches. If no member presents more than two of the four speeches, in how many different orders could the members give speeches?
(A)720
(B)1080
(C)1170
(D)1470
(E)1560
Number of members who could give the first speech = 6.
Number of remaining members who could give the second speech = 5.
Number of remaining members who could give the third speech = 4.
Number of remaining members who could give the fourth speech = 3.
To combine the choices above, we multiply:
6*5*4*3 = 360.
3 different speakers:
One member must give a pair of speeches.
Number of pairs of speeches that can be formed from 4 choices = 4C2 = 6.
Number of members who could give the pair of speeches = 6.
Number of remaining members who could give one of the two remaining speeches = 5.
Number of remaining members who could give the one remaining speech = 4.
To combine the choices above, we multiply:
6*6*5*4 = 720.
2 different speakers:
The speeches can be divided as follows:
1,2....3,4
1,3....2,4
1,4....2,3
Number of ways to divide the speeches = 3.
Number of members who could give the first pair of speeches = 6.
Number of remaining members who could give the remaining pair of speeches = 5.
To combine the choices above, we multiply:
3*6*5 = 90.
Total number of orderings = 360+720+90 = 1170.
The correct answer is C.
