Question

A group of 10 friends decides to head up to a cabin in the woods (where nothing could possibly go wrong). Unfortunately, a number of these friends have dated each other in the past, and things are still a little awkward. To get to the cabin, they need to divide up into some number of cars, and no two people who dated should be in the same car.(a) What is the smallest number of cars you need if all the relation-ships were strictly heterosexual? Represent an example of such a situation with a graph. What kind of graph do you get?(b) Because a number of these friends dated there are also conflicts between friends of the same gender, listed below. Now what is the smallest number of conflict-free cars they could take to the cabin? Friend A B C D E F G H I J Conflicts CFJ J AEF H CFG ACEGI EFI D AFG B

Answer 1

Step-by-step explanation:

A) The smallest number kg cars needed if all the relationships were heterosexual will be 3.

From the file attached, you will see that A, C, F and I will be in one car, B, J, H and E in another Car while G and D will be in the third car. Making it 3 cars.

B) the number of conflict free cars is therefore 3.

C) And these figures are important in coloring because of different permutations and combinations.

Go through the attached file to get a better understanding of the explanation.

Answer 2

The smallest number of conflict-free cars they could take to the cabin is 3.

Step-by-step explanation:

A C

I F

B J

H E

G D

The minimum number of cars can be 3 .

A, C, F, and I all in car 1.

B, J, H, E are in car 2.

G and D are in car 3.