Permutations

baladon99 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

here three cases are possible:
1) when all four are different, (i.e. persons who are giving speeches are different).
2)1 person giving 2 speeches and rest two gives 1 speech each.
3) 2 persons giving 2 speeches each.

case 1) out of 6 persons 4 persons can be selected in 6C4 ways, and the selected 4 persons can give speeches in 4! ways, hence total no. of ways= 6C4*4!=15*24=360
case 2) out of 6 person 1 person who will give 2 speeches can be selected in 6C1, remaining 2 persons who will give 1 speech each can be selected in 5C2, now selected person can deliver speeches in 4!/2!;( here four speeches are x,x,y,z x deliver by same person and y,z deliver by the different persons, hence total no. of arrangements=4!/2!),therefore required arrangements=6C1*5C2*(4!/2!)=720

case 3) now out of 6 persons 2 persons who will deliver 2 speeches each can be selected in 6C2 ways, now the selected can deliver speeches in 4!/(2!*2!), therefore total no. of arrangements =
6C2*(4!)/(2!*2!)=90

hence required no. of arrangements=360+720+90=1170; hence C