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12 Days of Christmas GMAT Competition 2025 - Day 3: A tech conference

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forestmayank

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There are six speakers who can present in 6! ways, i.e. 720 different combinations.

But the restriction is that ABCD should give presentation in A-B-C-D order only.

These 4 can be arranged in 4! ways, i.e. 24 ways in which all four can be reshuffled, but out of these 24 different combinations, only 01 combination would be in the order A-B-C-D. Considering this,

720/24 = 30 ways in which A-B-C-D will be in this order with other speakers presenting anywhere on remaining 2 places.

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The four speakers are clubbed as one group. So, there are three positions. Hence, we have 3! ways of arranging them=6... Therefore, the answer is choice: A..

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18 Dec 2025, 05:31

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IMO - answer is A

Order does matter so it is permuatation. Lock the seats in the order and calculate the remaining. It gives 3! = 6

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18 Dec 2025, 05:35

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Expectedorder: A>B>C>D
Total formations of 6 people without constraints = 6! = 720
A B C D can be arranged in 4! ways of which only one will satisfy the criteria
Hence total ways satisfying criteria = 6!/4! = 30

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18 Dec 2025, 05:45

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Alex, Bruno, Carla, Denise, Speaker 5, Speaker 6

Alex<<Bruno
Bruno<<Carla
Carla<<Denise

Let treat [Alex, Bruno, Carla, Denise] as 1 speaker thus, in essence, we are arranging 3 speakers - [ABCD], Speaker 5, Speaker 6.
3 speakers can be arranged in 3! ways = 3x2x1 = 6

[ABCD] S5 S6
[ABCD] S6 S5
S5 [ABCD] S6
S6 [ABCD] S5
S5 S6 [ABCD]
S6 S5 [ABCD]

Number of ways Alex, Bruno, Carla and Denise can be arranged = 1

Hence, 6 speakers can be arranged in 3! ways.

Bunuel

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Sujithz001

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18 Dec 2025, 05:55

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6 time slots.

Only order constraint: A > B > C > D

Since ABCD order is fixed, consider this as 1 element,

so number of possible arrangements: 3! = 6.

Answer: 6

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A tech conference schedules 6 speakers, including Alex, Bruno, Carla, and Denise, to give presentations in 6 different time slots, one speaker per slot. If Alex must speak earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise, in how many different ways can the 6 speakers be scheduled?
Total number of speakers = 6
A<B, B<C, C<D
Let other two speakers are E and F.
Total ways of arranging 6 speakers =6!
Ways of arranging 4 speakers = 4!
Out of those 24 ways, according to given conditions they can only be arranged in 1 way.
Total valid arrangments =6!/4! =6*5= 30

D. 30

redandme21

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number of unrestricted permutations of all the 6 people = 6!

number of permutations of the 4 people (Alex, Bruno, Carla and Denise) = 4!

From these, only 1 permutation is in the correct order: Alex before Bruno before Carla before Denise.

In the 6! permutations, all the 4! permutations are evenly distributed, so:

6!/4! = 6*5 = 30

IMO D

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We have 6 slots: 1,2,3,4,5,6
Pick 4 distinct slots for A,B,C,D in 6C4=15 ways because, once slots are chosen, the order of A,B,C,D is fixed.

The remaining 2 slots can be pick in 2!=2 ways

15*2=30 ways

Answer D

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To select 4 slots out of 6 for 4 speakers A, B, C & D, with their order set (A before B before C before D), we can approach this part like a combination problem since there is only one arrangement.

6C4 = 15

To arrange the other 2 speakers, there are 2! ways.

Therefore, total arrangements = 15 x 2! = 30

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Total permutations of 6 speakers: 6!=720.

The restriction Alex<Bruno<Carla<Denise means of the 4!=24 possible orders of Alex, Bruno, Carla, and Denise only 1 is allowed.

720/24=30

The answer is D

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The correct answer is Option D

So, let's say we have 6 different people where the order doesn't matter so we can fill the places in 6! ways, but in here there is an order which applies to 4 persons, so the correct answer will be

6!/4! = 30

So the correct answer is 30

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The order must be Alex earlier than Bruno earlier than Carla earlier than Denise.

There are 6 positions that must be filled with 4 people in that order 6C4=15

The other 2 positions can be filled in 2!=2

Combining: 15*2=30

The correct answer is D

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Bunuel

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Total placements is 6! = 720

We need A,B,C,D to be in order 4! = 24 ways are possible but 1 satisfies the order ABCD (1/24 ways)

= 720 * (1/24)
= 30 ways are possible

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The problem can be solved with 2 methods:

1. Step by step count:
=> Total permutations of 6 speakers => 6! = 720
Now, as four places are constrained in one order - they give 4! different relative orders when not constrained. Hence,
The valid fractions for the permutations are 1/4! = 1/24
=> Correct Answer = 720/24 = 30

2. Partition method:
The order given is :
|Alex|Bruno|Carla|Denise|
Consider the partitions as places.

Let the other two people be X and Y. First, X can be placed in the place of any one of the partitions i.e. 5 ways.

Lets say that one of the ways is:
|X|Alex|Bruno|Carla|Denise|

Now, Y can take any one of the partitions i.e. 6 ways.
Total ways = 5*6 = 30

Hence the correct answer = 30 => D

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The question says _A_B_C_D_ Basically A>B>C>D
Now we can place E & F in 5 gaps which can be done in 5C2 ways= 10 and E&F can be interchanged= 10*2=20 Ways
No suppose E&F can be placed together at any Gap, then same can be done in 5C1 ways and E&F cane be interchanged= 5*2= 10 Ways

Thus total possible options= 30
Answer is D

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the answer is option e 120

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6! total ways

half of those A is before B -> so divide by 2
a third of those, C is after A and B -> so divide by 3
a fourth of those, D is after A, B, and C -> so divide by 4

you end up with 6!/4! = 30

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Hi Bunuel, could you please explain the first approach? Why did we divide by 4!? What would have been the answer if there were 2 cases instead of 1 for ABCD? Just to understand the logic of the approach.

Bunuel

The total number of ways of scheduling 6 speakers without constraints is 6!.

We want a sequence where Alex speaks earlier than Bruno, Bruno earlier than Carla, and Carla earlier than Denise. That is just one specific relative order out of the 4! possible relative orders of Alex, Bruno, Carla, and Denise within the total 6! schedules. So the answer is 6!/4! = 30.

Alternatively, choose 4 positions out of 6 for the four speakers in the required order A - B - C - D: 6C4 = 15. Then arrange the remaining two speakers in the remaining two slots in 2! ways, so 15 * 2! = 30.

Answer: D

Bunuel

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Gmat232323

We divide by 4! because, across all 6! schedules, the four people Alex, Bruno, Carla, Denise appear in all 4! relative orders equally often. Only 1 of those orders is A before B before C before D, so we keep 1 out of 4! (1/4!) and get 1/4! * 6! = 6!/4!.

If instead there were 2 acceptable relative orders for those four (for example, two different allowed internal orders), then you would keep 2 out of 4!, so the count would be 2/4! * 6! = 2 * 6!/4!.