Question

.В группе из 25 человек любые двое имеют общего знакомого. Докажите, что из этой группы можно не менее, чем 36 способами выбрать пару знакомых школьников. Даю 30 баллов

Answer 1

Proof that there are at least 36 ways to choose a pair of familiar schoolchildren from a group of 25

To prove that there are at least 36 ways to choose a pair of familiar schoolchildren from a group of 25, we can use the principle of the pigeonhole principle. The pigeonhole principle states that if you have more pigeons than pigeonholes, then at least one pigeonhole must contain more than one pigeon.

In this case, the "pigeons" are the pairs of familiar schoolchildren, and the "pigeonholes" are the possible pairs that can be formed from the group of 25. Since any two people in the group have a common acquaintance, we can consider each person as a pigeon and each pair of people as a pigeonhole.

Given that there are 25 people in the group, we can calculate the number of possible pairs using the combination formula:

Number of possible pairs = C(25, 2) = 25! / (2! * (25-2)!) = 25 * 24 / 2 = 300

Therefore, there are 300 possible pairs that can be formed from the group of 25.

Now, let's assume that there are fewer than 36 ways to choose a pair of familiar schoolchildren. This means that there are 35 or fewer pairs of familiar schoolchildren.

Using the pigeonhole principle, if there are 35 or fewer pairs of familiar schoolchildren, and we have 300 possible pairs, then there must be at least one pair that is not familiar with each other. This contradicts the given statement that any two people in the group have a common acquaintance.

Therefore, our assumption that there are fewer than 36 ways to choose a pair of familiar schoolchildren is false. Hence, there must be at least 36 ways to choose a pair of familiar schoolchildren from a group of 25.

Therefore, we have proven that from a group of 25 people, there are at least 36 ways to choose a pair of familiar schoolchildren.