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Over the course of a full-day seminar, 66 students must give short pre

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Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   Updated on: 08 Feb 2015, 04:11
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I not able to understand the solution on the Mangoosh blog ..

Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!

Link : https://magoosh.com/gmat/2015/counting-practice-problems-for-the-gmat/
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Originally posted by shelrod007 on 07 Feb 2015, 19:46.
Last edited by Bunuel on 08 Feb 2015, 04:11, edited 1 time in total.
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Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   22 Feb 2015, 13:31
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icetray wrote:
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ok shelrod, we will solve it in other way..
lets see how many ways three places can be choosen for this three people out of 66 places.. it is 66C3... here we take combination and not permutation because we are taking selection and not arrangement of these three speakers..
remaining 63 can be arranged in 63!, and here arrangement makes a difference..
so ans becomes 66C3*63!=63!*66!/(63!*3*)=66!/3!...
yes we div by3! because only one arrangement out of 3! of RSM is valid..


Why are we selecting 66C3 ? Won't this give all the combinations of RSM? I thought that there were constraints on RSM such that R has to go before S, before M?
I understand that the "other 63" can be arranged with 66! but I don't understand how 66C3 would satisfy the RSM conditions... Does 66C3 not give "the number of combinations that 3 people (RSM) can be chosen out of 66 people", which would mean that RSM, SRM, MSR etc. would be included within 66C3 ?

Thanks


Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!

66 presentations can be ordered in 66! ways. We want only those sequences which have Ruth - Sam - Matt ordering in them.

Ruth's, Sam's and Matt's presentations can be ordered in 3! = 6 ways in those 66! cases: RSM, RMS, MRS, MSR, SRM and SMR. So, 1/6th of 66! would have RSM sequence in them, 1/6th of 66! would have RMS sequence in them, and so on. We need only RSM sequence, which as we deduced would appear on 1/6th of 66!, so the answer is 66!/6.

Answer: E.
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   07 Feb 2015, 20:24
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shelrod007 wrote:
I not able to understand the solution on the Mangoosh blog ..

Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!

Link : https://magoosh.com/gmat/2015/counting-practice-problems-for-the-gmat/


hi shelrod..
lets take a smaller value , 4 A, B, C and D.. there are 4 people in which there is a requirement of A speaking before B and rest can come anywhere..
so it can be ..A..B.. but not ..B..A..
now if we fix two others say at 1st and 2nd.. we will have C,D,A,B and C,D,B,A, out of these only one is correct.. similarily all other scenarios will have one case where A will be befor B and one wher B before A..so only one will be correct..

in this Q, there are three people where restrictions have been put.. R>S>M...rest can be anywhere..
if u take a fixed position for remaining 63, these three people can be placed in 3! ways amongst each other. out of these 3!, only one is correct where r>s>m..
similarily for all possible 66! ways, in which 63 people can be fixed and these 3 people can be fixed only in 1 way out of 3! possible ways..
this is the reason we divide 66! by 3! to cater for the restriction put..
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   07 Feb 2015, 21:17
chetan2u

I understood RSM can be arranged in 3 ! ways among the 63 people . Out of which only 1 order would be correct .

So total would be 66 ! ways in which 3 ! are the ways RSM is arranged .

But here i am confused about the division is it divided because out of the 66 ! ways only 1 is the correct arrangement for RSM ?
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   07 Feb 2015, 21:32
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shelrod007 wrote:
chetan2u

I understood RSM can be arranged in 3 ! ways among the 63 people . Out of which only 1 order would be correct .

So total would be 66 ! ways in which 3 ! are the ways RSM is arranged .

But here i am confused about the division is it divided because out of the 66 ! ways only 1 is the correct arrangement for RSM ?


ok shelrod, we will solve it in other way..
lets see how many ways three places can be choosen for this three people out of 66 places.. it is 66C3... here we take combination and not permutation because we are taking selection and not arrangement of these three speakers..
remaining 63 can be arranged in 63!, and here arrangement makes a difference..
so ans becomes 66C3*63!=63!*66!/(63!*3*)=66!/3!...
yes we div by3! because only one arrangement out of 3! of RSM is valid..
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   16 Feb 2015, 01:30
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shelrod007 wrote:
I not able to understand the solution on the Mangoosh blog ..

Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!

Link : https://magoosh.com/gmat/2015/counting-practice-problems-for-the-gmat/


66! are the total number of ways . Total number of ways to arrange R,S,M = 3!= 6 , clearly one 6th of these arrangements will have RSM , other 1/6th will have RMS and so on ..

so the answer is 66!/6
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   22 Feb 2015, 12:21
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ok shelrod, we will solve it in other way..
lets see how many ways three places can be choosen for this three people out of 66 places.. it is 66C3... here we take combination and not permutation because we are taking selection and not arrangement of these three speakers..
remaining 63 can be arranged in 63!, and here arrangement makes a difference..
so ans becomes 66C3*63!=63!*66!/(63!*3*)=66!/3!...
yes we div by3! because only one arrangement out of 3! of RSM is valid..


Why are we selecting 66C3 ? Won't this give all the combinations of RSM? I thought that there were constraints on RSM such that R has to go before S, before M?
I understand that the "other 63" can be arranged with 66! but I don't understand how 66C3 would satisfy the RSM conditions... Does 66C3 not give "the number of combinations that 3 people (RSM) can be chosen out of 66 people", which would mean that RSM, SRM, MSR etc. would be included within 66C3 ?

Thanks
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   22 Feb 2015, 13:42
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shelrod007 wrote:
I not able to understand the solution on the Mangoosh blog ..

Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!

Link : https://magoosh.com/gmat/2015/counting-practice-problems-for-the-gmat/


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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   22 Feb 2015, 13:43
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Bunuel wrote:
shelrod007 wrote:
I not able to understand the solution on the Mangoosh blog ..

Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!

Link : https://magoosh.com/gmat/2015/counting-practice-problems-for-the-gmat/


Similar questions to practice:
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mother-mary-comes-to-me-86407.html (or: mary-and-joe-126407.html);
six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   07 May 2016, 06:03
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shelrod007 wrote:
I not able to understand the solution on the Mangoosh blog ..

Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!

(B) 63!

(C) 63!/3!

(D) 66!/3

(E) 66!/3!



ANOTHER APPROACH

first fix those 3 students in the required order

_ R_ S _ M _

Now we can choose 1 place out of 4 for the next student (say A) i.e 4C1 ways

_ A _ R _ S _ M _

For 5th student, we can choose from 5 places i.e 5C1 ways

Similiarly , for 66th student 66C1 ways

Therefore, total ways = 4C1 x 5C1 x......x66C1 = 66!/3!
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   15 Jul 2021, 17:27
Bunuel wrote:
icetray wrote:
Quote:
ok shelrod, we will solve it in other way..
lets see how many ways three places can be choosen for this three people out of 66 places.. it is 66C3... here we take combination and not permutation because we are taking selection and not arrangement of these three speakers..
remaining 63 can be arranged in 63!, and here arrangement makes a difference..
so ans becomes 66C3*63!=63!*66!/(63!*3*)=66!/3!...
yes we div by3! because only one arrangement out of 3! of RSM is valid..


Why are we selecting 66C3 ? Won't this give all the combinations of RSM? I thought that there were constraints on RSM such that R has to go before S, before M?
I understand that the "other 63" can be arranged with 66! but I don't understand how 66C3 would satisfy the RSM conditions... Does 66C3 not give "the number of combinations that 3 people (RSM) can be chosen out of 66 people", which would mean that RSM, SRM, MSR etc. would be included within 66C3 ?

Thanks


Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!

66 presentations can be ordered in 66! ways. We want only those sequences which have Ruth - Sam - Matt ordering in them.

Ruth's, Sam's and Matt's presentations can be ordered in 3! = 6 ways in those 66! cases: RSM, RMS, MRS, MSR, SRM and SMR. So, 1/6th of 66! would have RSM sequence in them, 1/6th of 66! would have RMS sequence in them, and so on. We need only RSM sequence, which as we deduced would appear on 1/6th of 66!, so the answer is 66!/6.

Answer: D.


HI Bunuel

While I do understand this approach, the answer choice D is 66!/3 (and not 6) and the correct answer appears to be 66!/3!
What is it that we can amend in the above answer to get to the right answer?
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   16 Jul 2021, 00:26
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sm1510 wrote:
Bunuel wrote:
icetray wrote:

Why are we selecting 66C3 ? Won't this give all the combinations of RSM? I thought that there were constraints on RSM such that R has to go before S, before M?
I understand that the "other 63" can be arranged with 66! but I don't understand how 66C3 would satisfy the RSM conditions... Does 66C3 not give "the number of combinations that 3 people (RSM) can be chosen out of 66 people", which would mean that RSM, SRM, MSR etc. would be included within 66C3 ?

Thanks


Over the course of a full-day seminar, 66 students must give short presentations. Most of the presentations are independent, but Sam’s presentation refers to something in Ruth’s, so Ruth must go before Sam; similarly, Matt’s presentation refers to something in Ruth’s and something in Sam’s, so Matt’s must come after both of those. These three presentations need not be consecutive with each other: whatever the order is, Ruth’s must come before the other two but not necessarily first of the 66, and sometime later, Matt must come after the other two but not necessarily last of all. The other 63 participants can present in any order. How many orders obey these constraints?

(A) 22!
(B) 63!
(C) 63!/3!
(D) 66!/3
(E) 66!/3!

66 presentations can be ordered in 66! ways. We want only those sequences which have Ruth - Sam - Matt ordering in them.

Ruth's, Sam's and Matt's presentations can be ordered in 3! = 6 ways in those 66! cases: RSM, RMS, MRS, MSR, SRM and SMR. So, 1/6th of 66! would have RSM sequence in them, 1/6th of 66! would have RMS sequence in them, and so on. We need only RSM sequence, which as we deduced would appear on 1/6th of 66!, so the answer is 66!/6.

Answer: D.


HI Bunuel

While I do understand this approach, the answer choice D is 66!/3 (and not 6) and the correct answer appears to be 66!/3!
What is it that we can amend in the above answer to get to the right answer?


D was a typo there. The correct answer is E, as show in the solution.
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   24 Jul 2021, 00:52
The key to solving the problem is getting the right combination
since we can arrange each of them in 66! ways
However with the given constraint
Ritu first before sam and matt
matt after ritu and sam
there is only one possible possiblity that being
ritu sam and matt
they can be arranged in 3!
which we cannot have therefore this is divided
Hence IMO 66!/3--> E
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Re: Over the course of a full-day seminar, 66 students must give short pre [#permalink] New post   20 Jul 2023, 04:35
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