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Five friends A, B, C, D, and E go on a trip together. Each of them likes at least one of the following three activities: Countryside Sightseeing, Shopping, and Adventure Sports such that:

  • No two friends like exactly the same set of activities.
  • A, B and E like exactly two activities each.
  • D likes only Countryside Sightseeing.
  • C likes more number of activities compared to D.

The number of ways in which a pair of friends can be chosen for a trip from those who like Adventure Sports is ___

Entered answer:

Solution

✅ Correct Answer: 3

D likes only Countryside Sightseeing, so D's set has 11 activity. C must like more, so C likes 22 or 33 activities.


A, B and E each like exactly 22 activities. The only 22 activity options are {CS,Sh}\{\text{CS}, \text{Sh}\}, {CS,AS}\{\text{CS}, \text{AS}\} and {Sh,AS}\{\text{Sh}, \text{AS}\}. Since no two friends share a set, A, B and E together take all three options in some order.


C cannot take another 22 activity set without repeating one, so C likes all three.


The filled table:

FriendCSShAS
D✓\checkmark−-−-
One of A/B/E✓\checkmark✓\checkmark−-
One of A/B/E✓\checkmark−-✓\checkmark
One of A/B/E−-✓\checkmark✓\checkmark
C✓\checkmark✓\checkmark✓\checkmark

From the table, look at the AS column and count the rows that show ✓\checkmark.


These are the second A/B/E row, the third A/B/E row, and C. So 33 friends like Adventure Sports.


Number of pairs from these 33 friends:

3C2=3×22×1=3{}^{3}C_{2} = \dfrac{3 \times 2}{2 \times 1} = 3

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