Last visit was: 14 Jul 2024, 11:01 It is currently 14 Jul 2024, 11:01
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

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

SORT BY:
Tags:
Show Tags
Hide Tags
Intern
Joined: 19 Apr 2010
Posts: 5
Own Kudos [?]: 44 [37]
Given Kudos: 0
Math Expert
Joined: 02 Sep 2009
Posts: 94342
Own Kudos [?]: 640759 [8]
Given Kudos: 85011
General Discussion
Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 129
Own Kudos [?]: 589 [0]
Given Kudos: 0
Location: Toronto
Intern
Joined: 19 Apr 2010
Posts: 5
Own Kudos [?]: 44 [0]
Given Kudos: 0
thx Guys, now it make sense.
Senior Manager
Joined: 09 Jul 2013
Status:1,750 Q's attempted and counting
Affiliations: University of Florida
Posts: 421
Own Kudos [?]: 3006 [2]
Given Kudos: 630
Location: United States (FL)
GMAT 1: 570 Q42 V28
GMAT 2: 610 Q44 V30
GMAT 3: 600 Q45 V29
GMAT 4: 590 Q35 V35
GPA: 3.45
WE:Accounting (Accounting)
Re: An oratorical society consists of six members, and at an upc [#permalink]
2
Kudos
This appears to be from the GMAT Hacks question of the day archive here is the Official explanation

- 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).

Intern
Joined: 12 Jul 2017
Posts: 30
Own Kudos [?]: 58 [0]
Given Kudos: 18
Re: An oratorical society consists of six members, and at an upc [#permalink]
Bunuel wrote:
ilaila wrote:
i came across this problem in a word document so i dont know it is official Q or not.

-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

There can be 3 cases:

1. ABCD - 4 different members giving 1 speech each --> $$P^4_6=360$$;

2. AABC - 1 member presents two speeches and other two members one speech each --> $$C^1_6$$ - # of ways to choose the one who will present two speeches, $$C^2_5$$ - # of ways to choose other two members who will present one speech each, $$\frac{4!}{2!}$$ - # of ways these members can give their speeches (eg. AABC, ABCA, BCAA, ... basically # of permutations of 4 letters AABC). So, for this case: $$C^1_6*C^2_5*\frac{4!}{2!}=720$$;

3. AABB - two members present two speeches each --> $$C^2_6$$ - # of ways to choose two members who will present two speeches each, $$\frac{4!}{2!2!}$$ - # of ways these members can give their speeches (eg. AABB, ABAB, BBAA, ... basically # of permutations of 4 letters AABB). So, for this case: $$C^2_6*\frac{4!}{2!2!}=90$$.

Total: $$360+720+90=1170$$.

Hope it helps.

Hi everyone,
Can someone explain to me where I'm going wrong with this approach:

6 members can have 4 speeches as follows:
1) any of the six can start then any of the five etc... we get 6*5*4*3=360 for 4 speeches made by 4 different persons.
2) any of the six can start then the same one has to make another one and then another person takes place and makes 2 speeches so we have: 6*1*5*1=30. This one can be arranged in 4!/(2!*2!)=6 ways. So the total is 30*6=180.
3) any of the six can start then he/she has to make another speech and then the two remaining speeches will be given by 2 different persons resulting in 6*1*5*4=120. This can be arranged in 4!/2!=12 ways totaling in 1440.
the total for 1, 2 & 3 is 1440+180+360=1980.

if #2 is divided by 2 and #3 by 2 i will get the correct answer but i can't understand why. Can someone please explain?
Thank you
VP
Joined: 10 Jul 2019
Posts: 1385
Own Kudos [?]: 577 [1]
Given Kudos: 1656
Re: An oratorical society consists of six members, and at an upc [#permalink]
1
Kudos
The Key Errors I made (and are worth pointing out)

(1) the question is looking for the Order of the MEMBERS, not the speeches themselves. So it is NOT 4! Each Time for the 4 Different Speeches. We need to re-arrange the Members for each different scenario

(2)Choosing which person will GIVE 2 SPEECHES ---- vs. ----- Choosing which person will GIVE 1 SPEECH: is 2 DISTINCT GROUPS that you are choosing for. Thus, you need to make 2 Selections followed by another.

Scenario 1: Person A --- gives 2 Speeches
Person B -----gives 2 Speeches

(1) In How many ways can we select the 2 People who will give 2 Speeches each?

Since this is NOT a Distinct Group (each person will give 2 Speeches) we can just use the Combinations Formula:

"6 choose 2" = 6! / (2!4!) = 15 Ways

AND

(2) For Each of the Different Combinations of groups, how many ways can we uniquely ARRANGE the People speaking?

1 Arrangement: A - then A - then B - then B

in this case, A never comes down, so it does not matter which of A's 2 Speeches he gives 1st. We only care about the ORDER of the members, not the speeches.

A-A-B-B

Can be Arranged in: 4! / (2!2!) = 6 Arrangements

15 * 6 = 90 Ways we can Order the Members if 2 Members each give 2 Speeches

Scenario 2: Person A --- gives 2 Speeches
Person B --- gives 1 Speech
Person C ---- gives 1 Speech

(1) Now, when we select which person gives 2 Speeches vs 1 Speech, we are choosing for 2 DISTINCT GROUPS. Therefore, we need to make 2 Successive Selections.

In How many ways can we choose the 1 person who will give 2 Speeches?
"6 choose 1" = 6 Ways

AND

In How many ways can we choose the 2 people who will each give 1 Speech?
"5 choose 2" = 5! / (2!3!) = 10Ways

(2) Given all the Different Combinations of Groups, for Each possible Group how many ways can we ARRANGE the Members for which Order they speak?

A - A - B - C ------ this is 1 Arrangement
A - B - A - C ------- this is 1 Arrangement (person A comes down, then Person B comes up, then person A comes back up)

4! / (2!) = 4*3 = 12 Arrangements

(6 * 10) * 12 = 720 Ways to Order the Members where 1 Members gives 2 Speeches and Another 2 Members EACH give 1 Speech

Scenario 3: We have 4 People go up and EACH PERSON gives 1 Speech

(1)In How many ways can we CHOOSE 4 People to give 1 Speech out of a Total of 6 members?

"6 choose 4" = 6! / (4!2!) = 15 Ways

AND

(2) How many ways can we ARRANGE the 4 Unique Members order on stage?

4! = 24 Arrangements b/c each person is UNIQUE

15 * 24 = 360 Ways

90 Ways + 720 Ways + 360 Ways = 1, 170 Total Ways

-C-
Intern
Joined: 31 Mar 2023
Posts: 2
Own Kudos [?]: 1 [0]
Given Kudos: 2
An oratorical society consists of six members, and at an upc [#permalink]
In case it's helpful. I think its much easier to figure this out in reverse.

Let's start with the total number of way's 6 people can give four speeches (repeating is allowed for all). That is simply $$6*6*6*6=1296$$ options.

What about the case 1 person presents all 6 speeches. That is simply $$_6C_1=6$$ options.

Next let's look at the case 1 person presents 3 speeches. That will be all variations of XXXY. We can select X in $$_6C_1=6$$ ways and Y in $$_5C_1=5$$ ways so 30 combinations of X and Y. In terms of possible orders for XXXY, this is simply 4 ($$\frac{4!}{3!}$$) orders. That means there are $$30*4=120$$ options.

Given, we want all speeches where 1 person doesn't do 3 or 4 speeches. The resulting options are $$1296-6-120=1170$$, choice (C).
An oratorical society consists of six members, and at an upc [#permalink]
Moderator:
Math Expert
94342 posts