I understand how you come to the total amount of ways (60) for the denominator, but how can you acknowledge that the two empty (E) seats are identical when calculating 4!/2! *2, but not acknowledge it when calculating 5!/2! = 60?.
keiraria wrote:
hello
3 persons - one couple and one single are seated at random in a row of 5 seats
what is the probability that the couple does not sit together
please explain your approach
best regards
Let's find the probability that a couple sits together (right next to each other) and subtract that value from 1.
Total # of ways 3 persons \(C_1\), \(C_2\) and \(S\) to be seated in a row of 5 seats is \(\frac{5!}{2!}=60\). Consider this, we are interested in arrangement of \(C_1, \ C_2, \ S, \ E, \ E\), so in arrangement of 5 letters out of which 2 E's are identical (E denotes an empty seat);
# of ways for a couple to sit together is \(\frac{4!}{2!}*2=24\). Consider a couple as a single unit: \(\{C_1,C_2\}, \ S, \ E, \ E\), so we have total of 4 units out of which 2 E's are identical, # of arrangement of these units is \(\frac{4!}{2!}\), but \(C_1\), \(C_2\) within their unit can be arranged in 2 ways (\(\{C_1,C_2\}\) or \(\{C_2,C_1\}\)), so total # of arrangement for this case is \(\frac{4!}{2!}*2=24\);
\(P=1-\frac{24}{60}=\frac{3}{5}\).
Answer: D.