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12 Jan 2026, 06:46
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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
A. 6
B. 12
C. 24
D. 30
E. 120
12 Jan 2026, 06:46
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Official Solution:
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
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 \(\frac{6!}{4!} = 30\).
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
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
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 \(\frac{6!}{4!} = 30\).
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
