An ice-cream store has exactly 6 flavors of ice-cream. Each of 6 friends likes exactly 4 flavors.
Is there guaranteed to be a flavor which at least 4 friends like?
Yes. Let's say that there is no flavor which at least 4 friends like. This means that for each flavor, at most 3 friends can like that flavor. This would allow a total of 6 flavors x 3 = 18 preferences, which is less than the number of total preferences (6 friends x 4 preferences = 24). Thus, it is always guaranteed that there will be a flavor at least 4 friends like.
I don't see how this works
Perhaps an easier way to understand it would be this -- assume friends 1, 2, 3 like ice creams A, B, C and friends 4, 5, 6 like ice creams D, E, F. We know that each friend has to like exactly four flavors, so we must assign each of them to one more ice cream. (Right now, three friends like each ice cream.) No matter how you place their choices, the number will exceed three (and thus, at least four) for at least one ice cream.
Therefore, there is always guaranteed to be a flavor that at least 4 friends like. This is an extension of the Pigeonhole Principle -- if there are 8 pigeons and 7 holes, one hole must have at least two pigeons.
I hope this answers your question :)
To answer this question, we need to examine the scenario and analyze the possibilities.
We know that there are 6 flavors of ice cream, and each of the 6 friends likes exactly 4 flavors. Therefore, we can assume that each friend likes a unique combination of 4 flavors.
If we imagine the 6 friends arranged in a row, we can determine the maximum number of unique combinations of 4 flavors that can be liked by the friends. Since each friend likes 4 flavors and there are 6 friends, the total number of unique combinations of 4 flavors is given by the formula 6 choose 4, denoted as 6C4.
To calculate 6C4, we can use the binomial coefficient formula: n! / (r!(n-r)!), where n is the total number of items and r is the number of items chosen at a time.
In this case, plugging in the values, we have:
6! / (4!(6-4)!) = 6! / (4!2!) = (6*5*4*3*2*1) / ((4*3*2*1)*(2*1)) = 15.
So, there are 15 unique combinations of 4 flavors that can be liked by the friends.
Now, let's consider the worst-case scenario, where each flavor is liked by exactly 3 friends. In this case, we have 6*3 = 18 total likes across all the flavors.
Since there are 15 unique combinations, we can conclude that there must be at least one flavor that is liked by at least 4 friends. This is because, if there were no such flavor, the maximum number of likes across all flavors would be 3*15 = 45, which is greater than the 18 likes in our worst-case scenario.
Therefore, it is guaranteed that there will be a flavor liked by at least 4 friends in this scenario.