A salesman is arranging his schedule for visiting each of the three towns a, b and c, twice. Let A be the event that first and the last visits are in a. Write down the elements of the sample space and the event A.
Answers
Answer:
Step-bThe salesman must visit three towns, twice each. Let’s just make up an example of a trip he could take and see what we learn. Say he likes to go in order. So he visits a then b then c. Then he does that again. So his visits go abc abc. We can organize this into a list like this (a, b, c, a, b, c). Now we have a form we can work with. That list is now a vector with six spots to put something and we have three things we could put in each spot. So we could count them. 3 options for the first spot, 3 for the second, the third, and so on. Multiplying them together for each spot we get 36.
But wait! Counting like that includes trips like (a, a, a, a, b, c). The salesman must only visit each city twice. So some of these options are invalid. So let’s try to count his possible trips a different way. He has to decide when to go to a. He could do aaxxxx, axaxxx, axxaxx, axxxax, and so on. Essentially he has to choose where in that list to put two visits to city a. There’s a way to count that. It’s 6 choose 2. Next he has to choose when to go the city b. Since a is already taking two spots the salesman has four options for where to place his visits to city b. That’s 4 choose 2. Then the last two spots get the visits to city c and he has no choices left to make. Multiplying those together gives 6C2*4C2=15*6=90 possible trips.
Answer:
abcbca, abccba, acbbca, acbcba
Step-by-step explanation:
Sample space is 36
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abccab
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abcabc
acbacb
acbabc
acbbac
acbbca
acbcab
acbcba
baccba
bacabc
bacacb
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bacbca
baccab
bcaabc
bcaacb
bcabac
bcabca
bcacab
bcacba
cababc
cabacb
cabbac
cabbca
cabcab
cabcba
cbaabc
cbaacb
cbabac
cbabca
cbacab
cbacba