Problem Solving

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

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

4 different speakers:
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.

Similar to what Mitch has done here,

My approach--

1 speech per person-- this way we need 4 people out of 6. 4 people can be chosen in 6C4 ways.
and 4 speeches can be distributed in 4! ways. Total number of ways - 24* 15.= 360


second case - 2 speech per person
For this we need 2 people. This can be done in 6C2 ways.
Now let's say the two person selected are A and B
4 speeches can be distributed between A and B in 6 ways I.e in 4C2 ways
Total comes out to be 90


Third and last case
One person delivers 2 speech and rest 2 are divided between 2 people
Total number of people required 3
They can be selected in 6C3 ways
Now let's say they are A, B and C
We need to choose a person who will deliver 2 speech
This can be done in 3 ways, say A
The other 2 left are B and C
Now, A can choose 2 speech from 4 in 4C2 ways
The other 2 can chose from 2 speech in 2 ways
Therefore total comes out to be 20*3*6*2 = 720

do the total of 3 case and we will get our answer,
Which is C

The subtraction approach works well here.

Total possibilities 6*6*6*6 =1296

Ways that 1 person gives all 4 speeches: 6

Ways that 1 person gives 3 speeches:
Who gives the 3 speeches: 6 options
Which 3 speeches does he give? 4C3 = 4 options
Who gives the last speech: 5 options

6*4*5 = 120

1296 - 120 - 6 = 1170.

Hi Mitch,

Thanks for the explanation.. But can you please explain how did u get this:

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.

Edited.

@Mitch

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.

This is the point I'm having trouble with. Assume the speakers are A B C D E F Now if I need two different speakers example AABB Then I can select two in 6C2= 15 Ways And the four speeches to be assigned to them can be done in 4C2=6 Ways. So possibilities=15*6=90

Is my reasoning correct?

gmatboost wrote:The subtraction approach works well here.

Total possibilities 6*6*6*6 =1296

Ways that 1 person gives all 4 speeches: 6

Ways that 1 person gives 3 speeches:
Who gives the 3 speeches: 6 options
Which 3 speeches does he give? 4C3 = 4 options
Who gives the last speech: 5 options

6*4*5 = 120

1296 - 120 - 6 = 1170.

Hi am unable to follow the part in green!

knight247 wrote:@Mitch

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.

This is the point I'm having trouble with. Assume the speakers are A B C D E F Now if I need two different speakers example AABB Then I can select two in 6C2= 15 Ways And the four speeches to be assigned to them can be done in 4C2=6 Ways. So possibilities=15*6=90

Is my reasoning correct?

Looks good.

In my solution:
I counted the number of ways the speeches could be divided without regard to order (4C2/2! = 3).
I took order into account by counting the number of ways the speakers could be arranged (6*5 = 30).

In your solution:
You counted the number of ways the speakers could be selected without regard to order (6C2 = 15).
You took order into account by counting the number of ways the speeches could be assigned (4C2 * 2C2 = 6).

Either way is fine.

I understand how to arrange 6 people into 4 individual speeches (6! over 2! = 360).

I falter at the next step. If 3 people are speaking, (3 people speak once, 1 person gives a pair of speeches), then...??? What I did was run 6!/3!*3! to get 20 different groups of 3 speaks, then multiplied by 4!, then number of orders of speeches those 20 groups could speak in = 480.

Likewise I got 15 different groups of 2 speakers, multiplied by 4! for the order those speeches could go in, for 360. 360+360+480 = 1200. (frown face)

Also, just to clarify, in forum speak, 4C2 means 4!/2! correct?