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

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

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New post 25 Jun 2010, 03:11
2
6
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

36% (02:23) correct 64% (02:20) wrong based on 95 sessions

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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

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?
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Re: Comb problem!?  [#permalink]

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New post 25 Jun 2010, 04:06
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: \(360+720+90=1170\).

Answer: C.

Hope it helps.
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Re: Comb problem!?  [#permalink]

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New post 25 Jun 2010, 11:48
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

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)


Hi!

You're missing some options because you've treated all 3 people the same.

You need to remember that of the 3 people, 1 is going twice and two are going once. Since one of the 3 is going once, there are 3C1 possibilities for the double speaker.

So, to convert your work to the correct answer, you need to multiply:

240 * 3C1 = 240 * 3 = 720.

Now your solution matches the correct answer.
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Re: Comb problem!?  [#permalink]

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New post 26 Jun 2010, 08:56
thx Guys, now it make sense.
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Re: An oratorical society consists of six members, and at an upc  [#permalink]

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New post 17 Nov 2013, 21:15
1
This appears to be from the GMAT Hacks question of the day archive here is the Official explanation

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).

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Re: An oratorical society consists of six members, and at an upc  [#permalink]

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New post 16 Mar 2018, 03:49
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: \(360+720+90=1170\).

Answer: C.

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
Re: An oratorical society consists of six members, and at an upc &nbs [#permalink] 16 Mar 2018, 03:49
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