<p>I searched the tagged thread for answers, but it was not there.</p>
<p>p. 392 number 10. </p>
<p>There are 3 roads from Plattsville to Ocean Heights and 4 roads from Ocean Highs to Bay Cove. If Martina drives from Plattsville to Bay Cove and back, passes through Ocean Heights in both directions, and does not travel any road twice, how many different routes for the trip are possible?</p>
<p>A. 72 (right answer)
B. 36
C. 24
D. 18
E. 12</p>
<p>I was able to figure the answer out, but only by drawing a simple diagram and thinking about it.</p>
<p>Does anybody if one can solve this by Permutations or Combinations? If so, could you please explain. </p>
<p>Or is the only way to solve it by thinking it out?</p>
<p>Let’s label the roads:
3 roads from Plattsville to Ocean Heights (and back) - 1, 2, and 3
(numbers #);
4 roads from Ocean Highs to Bay Cove (and back) - A, B, C, D, and E
(letters L).</p>
<p>We are looking for all possible arrangements
(#1, L1, L2, #2) where both #'s and L’s don’t repeat (Martina does not use
the same road twice).
There are
3 choices for the first number #1,
4 choices for the first letter L1,
3 choices for the second letter L2,
2 choices for the second number #2.
According to Fundamental Counting Principle a total number of arrangements
is 3 x 4 x 3 x 2 = 72.</p>
<p>If you really badly want to “solve this by Permutations or Combinations”,
you need to use Permutations formula, because the order of terms
in an arrangement is important: AB is different from BA.</p>
<p>We’ll have to rewrite
(#1, L1, L2, #2) as
(#1, #2, L1, L2) - it has the same number of arrangements, but now similar
terms are next to each other.
(#1, #2) can be arranged in 3P2 = 6 ways
(L1, L2) can be arranged in 4P2 = 12 ways.
Total = 6 x 12 = 72.</p>