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

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Bunuel

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17 Dec 2025, 10:00

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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

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paragw

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17 Dec 2025, 09:09

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There are 6 speakers, so without restrictions they can be scheduled in 6! ways.

Alex before Bruno before Carlo before Denise.
Among themselves they can be arranged in only 1 valid order instead of 4!.

So divide the total arrangements by 4!:

6!/4!
=> (6*5)
=> 30

Correct answer: D

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17 Dec 2025, 09:12

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let us consider [abcd] be one block and remaining is r1 and r2
[abcd]r1,r2
we will treat them as 3!=6
[abcd] cannot change

ManifestDreamMBA

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17 Dec 2025, 09:18

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Let all the speakers be A,B,C,D,E and F

They can be arranged in 6! ways but the order for A<B<C<D is fixed so the arrangement among these 4 need to be removed
They can be arranged in 4! ways

Total ways for scheduling = 6!/4! = 6*5 = 30

Answer D

Bunuel

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17 Dec 2025, 09:21

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Bunuel

Deconstructing the Question
Total speakers: 6. Let the speakers be A, B, C, D, and two others (let's call them E and F).
Constraint: A must be earlier than B, B earlier than C, and C earlier than D.
This establishes a fixed relative order: $A \rightarrow B \rightarrow C \rightarrow D$.
Method 1: The Slot Selection Approach
1. We have 6 time slots available: $\_ \_ \_ \_ \_ \_$
2. First, place the 4 constrained speakers (A, B, C, D). We need to choose 4 slots out of the 6 available for them.
Since their relative order is fixed (A is 1st, B is 2nd, etc.), once we pick the 4 slots, there is only 1 way to arrange A, B, C, and D into those specific slots.

Calculation:
$^{6}C_{4} = \frac{6 \times 5}{2 \times 1} = 15$ ways to choose the slots.

3. Now we have 2 empty slots remaining and 2 unconstrained speakers (E and F).
We can arrange E and F in the remaining slots in $2!$ ways.

Calculation: $2 \times 1 = 2$ ways. 4. Total Arrangements = (Ways to place A,B,C,D) $\times$ (Ways to place E,F)

Total = $15 \times 2 = 30$.

Method 2: The Probability/Permutation Approach
Total permutations of 6 distinct speakers without restrictions = $6! = 720$.
Consider the subset of 4 speakers {A, B, C, D}.
In any random permutation, these 4 speakers can be arranged among themselves in $4! = 24$ different ways.
Only 1 of these 24 ways follows the specific order $A \rightarrow B \rightarrow C \rightarrow D$.
Therefore, only $\frac{1}{24}$ of the total permutations are valid.

Calculation: $\frac{Total Permutations}{Ways-to-arrange-the-constrained-group} = \frac{6!}{4!} = \frac{720}{24} = 30$.

Answer: D

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17 Dec 2025, 09:29

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Total ways of arranging 6 people = 6!
Now as A, B, C, D have a specific order we have to divide by 4!.

So final answer = 6!/4! = 30 (Option D)

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17 Dec 2025, 09:30

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Total number of ways 6 people can be be arranged in 6 slots: 6!

Conditions: A before B before B before C before D.
This can be done only in 4! ways.

Total possible ways : (6!)/(4!) = 30.
Ans: C.

Bunuel

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17 Dec 2025, 09:34

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There are 6 total speakers, 1 per slot -> the total possible arrangements are 6! = 720.
But there is only 1 way to arrange the 4 speakers ABCD out of the possible 4!=24 arrangements. -> 1/24 schedules work with the restrictions

total arrangements = 720
valid arrangements = 720 * (1/24) = 30

720/24 = 30
D

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17 Dec 2025, 09:35

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Let E and F be other two speakers.
for E, there are 5 possible options:(represented by _):
_A_B_C_D_
Similarly, for F there are 5 such options.
In all there are 5*5 = 25 options for E and F

Now, EF can come together, again in 5 ways in places: _A_B_C_D_.
Therefore, these 25 options can be divided in tow groups= 20+5, 5 being when they are together.
IN these 5 options, by swapping EF by FE, we get 5 more options. Therefore total we have 20+5+5= 30 options

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17 Dec 2025, 09:40

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The restricted order: Alex < Bruno < Carla < Denise
There are 6 speakers, so the 2 remaining speakers can be alloted for 5 different places: _ Alex _ Bruno _ Carla _ Denise _ ( _ = 1 place)
Case 1: 2 speakers can take one _ place => there are 5*2 = 10 ways
Case 2: 2 speakers can take 2 _ places => there are 5P2 = 20 ways (because the slots must follow order so we use permutation, not combination)
Total = 10+20 = 30 ways

Answer: D

Bunuel

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17 Dec 2025, 09:40

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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?

A. 6
B. 12
C. 24
D. 30
E. 120

total speakers are 6
ABCDEF
one speaker per lot and among ABCD the sequence is set
Let ABCD be X
XEF ; 6 ways
similarly we can have ways
in between ABCD we can place E & F separately in 5 ways each total 10
EF when together can be placed between ABCD in again 5 ways
FE when together can be place between ABCD in 5 ways
in between each ABCD EF can be placed EAFBCD ; AEBFCD; ABECFD; ABCEDF; 4 ways
total possible ways
6+10+5+5+4 = 30
OPTION D , 30 is correct

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17 Dec 2025, 09:42

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Let the sequence be from early to late

Alex < Bruno < Carla < Denise

Total number of ways to arrange 6 speakers without any condition = 6!

The number of ways 4 speakers Alex, Bruno, Carla & Denise can be arrange without any condition = 4!

Since the sequence for the 4 speakers is fixed, total number of ways to arrange 6 speaker in which 4 speakers have a particular sequence = 6!/4! = 30

IMO D

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17 Dec 2025, 09:47

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Total number of ways in which 6 speakers can be presented will be 6! = 720 ways.
Given that A,B,C,D should be arranged in such a manner that A should come before B, B should come before C and C should come before D. No. of ways to arrange these 4 speakers is 4! =24 and only 1 out of 24 arrangements can meet the above given conditions.
So, total no. ways which will meet above condition will be 720/24 = 30. (D)

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17 Dec 2025, 09:48

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Bunuel

Case 1 : The other two speaks don't speak immediately after each other

_ A _ B _ C _ D _

Select two space from the available spaces in 5C2 ways and each of the selected space can be arranged in two ways

5C2 * 2 = 20

Case 2: The speakers speak after one another

_ _ A _ _ B _ _ C _ _ D _ _

One of the two blocks can be selected in 5 ways and each blocked can be arranged in two ways

Total = 10

Grand Total = 20 + 10 = 30

Option D

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17 Dec 2025, 09:51

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we have 6 diff. speakers with 6 slots----6 factorial = 720
given cond. alex<Bruni<carla<denise---1/4!

total permutation 6!/4! ----720/24=30

Bunuel

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17 Dec 2025, 09:54

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Bunuel

We have 6 slots, we need to pick 4 out of 6 slots, the number of ways to do the same is = 6C4 = 6!/(2!*4!) = 3*5 =15

In 4 slots picked, 4 slots can be assigned in the given order to A B C & D

In remaining 2 slots other two people can be arrange in 2! ways that is 4..

Total Ways = 15* 2= 30 D Answer

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17 Dec 2025, 09:55

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6 can appear in 6! ways = 720 ways
4 can appear in 4! ways =24 ways
Out of these 24, there is only one where Alex<Bruno<Carla<Denise
Answer = 720/24=30 ways

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17 Dec 2025, 10:03

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We want to keep the position of A, B, C and D fixed.
So out of the six slots, we pick any 4, and then in the arrange the remaining two people in the 2 slots left.

6C2 * 2 = 30
Option D.

Bunuel