Compute Probabilities Involving Permutations and Combinations
-Amy, Jean, Keith, Tom, Susan, and Dave have all been invited to a birthday party. They arrive randomly and each person arrives at a different time. In how many ways can they arrive? In how many ways can Jean arrive first and Keith last? Find the probability that Jean will arrive first and Keith will arrive last.
A) $720 ; 24 ; \frac { 1 } { 30 }$
B) $720 ; 15 ; \frac { 1 } { 48 }$
C) $120 ; 6 ; \frac { 1 } { 20 }$
D) $120 ; 10 ; \frac { 1 } { 12 }$

Question

Quizplus · Accepted Answer

The total number of ways they can arrive is $6!$ because there are 6 people and each has a unique position in the sequence of arrivals, which equals $720$. For Jean to arrive first and Keith last, the other 4 can arrive in any order, which is $4!$ ways, equaling $24$. The probability is the number of favorable outcomes ($24$) over the total number of outcomes ($720$), which simplifies to $\frac{1}{30}$.

Compute Probabilities Involving Permutations and Combinations -Amy, Jean, Keith, Tom 720;24;130720 ; 24 ; \frac { 1 } { 30 }720;24;301​

Compute Probabilities Involving Permutations and Combinations
-Amy, Jean, Keith, Tom $720 ; 24 ; \frac { 1 } { 30 }$