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bhanu29
Joined: 02 Oct 2024
Last visit: 27 Apr 2026
Posts: 363
Given Kudos: 263
Location: India
Schools: Anderson '27 Booth '27 Foster '27 Haas '27 Kellogg '27 Kellogg Mashall '27 McCombs '27 Ross '27 Scheller '27 Tepper '27 Yale '27 Sloan (MIT) Stanford '27
GMAT Focus 1: 675 Q87 V85 DI79 
GMAT Focus 2: 715 Q87 V84 DI86
GPA: 9.11
WE:Engineering (Technology)
Products:
Schools: Anderson '27 Booth '27 Foster '27 Haas '27 Kellogg '27 Kellogg Mashall '27 McCombs '27 Ross '27 Scheller '27 Tepper '27 Yale '27 Sloan (MIT) Stanford '27
GMAT Focus 2: 715 Q87 V84 DI86
Posts: 363
17 Dec 2025, 20:07
Kudos
Bookmarks
Bunuel
We have 6 slots and we need to preserve order ABCD
So first step choose any 4 slots among 6 to put ABCD in same order
6C4 = 6*5/2 = 15
Now put remaining 2 speakers in other slots and they can be rearranged in 2! ways
So total arrangements 15 * 2 = 30
Correct Answer: D
17 Dec 2025, 21:12
Kudos
Bookmarks
ordered 4 for 6 seats: 6C4 * 1(order A < B < C < D) = 15
remaining 2 speakers for 2 slots = 2!
total ways: 15*2 = 30
remaining 2 speakers for 2 slots = 2!
total ways: 15*2 = 30
17 Dec 2025, 21:25
Kudos
Bookmarks
If six people are to be arranged for 6 different time slots, we can prepare the schedule in 6! ways
But given the condition A<B<C<D, let us see how to accomodate this condition
If we were to arrange A B C D in different ways, we get 4! ways. But bringing in this condtion, we can only arrange them as A B C D in the order. All the other arrangement (eg: BCDA CDBA etc) do not hold good.
So we can take only 1 way to arrange A B C D from 4! ways
So, from arrangement of 6 of them, we have to divide it by 4!
Total number of ways the speakers can be scheduled given the condition = 6!/4! = 5*6 = 30
But given the condition A<B<C<D, let us see how to accomodate this condition
If we were to arrange A B C D in different ways, we get 4! ways. But bringing in this condtion, we can only arrange them as A B C D in the order. All the other arrangement (eg: BCDA CDBA etc) do not hold good.
So we can take only 1 way to arrange A B C D from 4! ways
So, from arrangement of 6 of them, we have to divide it by 4!
Total number of ways the speakers can be scheduled given the condition = 6!/4! = 5*6 = 30
Bunuel
