An oratorical society consists of six members, and at an upc

their answer was
Answer: C
This is a tricky one. There are several possible ways for the four speeches to be chosen. The first, and the simplest, is to have each of four members deliver one speech each. That's a more traditional permutations problem. Any of 6 speakers can be chosen for the first speech; any of the remaining five for the second speech; 4 for the third, and 3 for the second. Multiply those for the number of permutations of this sort:
(6)(5)(4)(3) = 360
Keep that number handy, because we'll need to add it to the results from the other possibilities.

The other "simple" possibility is that two members are chosen to give two speeches each. There are three ways to arrange two speakers:
XXYY (one member gives the first two speeches, another gives the last two)
XYYX (one member gives the first and last speeches, another gives the middle two)
XYXY (the members alternate)
For any of those, the number of possible permutations is 6 (the number of possible speakers designed "X" in either of the arrangements) multiplied by 5 (the number of possible speakers designed "Y", since one has already been chosen to be "X"). That is, there are 30 possible XXYY's, and 30 more possible XYYX's. That's 90 more, for a running total of 420.

Now for the tough part. The schedule could consist of one member who gives two speeches and two other members who give one speech each. Here are the possible ways schedules that include a setup like that:
XXYZ
XYXZ
XYZX
YXXZ
YXZX
YZXX
There are six arrangements that place two speeches by the same person among a schedule of four speeches. (By the way, this is equivalent to a combinations problem where we're choosing 2 from a group of 4.)

In each of those six arrangements, we're choosing 3 different members. The number of choices of members is (6)(5)(4) = 120. That's the number of permutations for any one of the six arrangements listed above. Thus, the total number of permutations for all of the different arrangements of this sort is (120)(6) = 720.

Finally, we have the following:
360 permutations with 4 speakers
90 permutations with 2 speakers
720 permutations with 3 speakers
Total: 360 + 90 + 720 = 1170, choice (C).

i cracked the question differently but got different answer!?
my approach was i can select any member regardless of order then check the arrangements for the speeches ( two different events) as below:

1- one member one speech:
first we need to select any 4 out of 6 ( arrangement dose not matter here) so,
6C4=6!/4!2!=6X5/2=15 --------1
now how many ways we can arrange these 4(XYZW) to make the speech so,
4P4=4!=24-------------2
now 15x24 = 360 ( same number they got)

2- two members four speeches
same approach
6C2=15---------------1
XXYY can be arranged as distinctive Permutations
4P4/2!2! = 6------------2
now 15x6=90 ( same number they got)

3- one member two speeches and two one each
6C3=20
XXYZ can be arranged as distinctive Permutations
4P4/2!=12
now 20x12= 240 ( not the same)

total is ( 360+90+240=690) it is not even an option?

which one is correct? i think they used Permutations for selecting members(why) and they cut the speech arrangements by half(not clear)?

any ideas?