Question

В каждом из 20 пеналов лежит либо 4, либо 5, либо 6 карандашей. а)Верно ли что найдется хотя бы 7 пеналов с одинаковым количеством карандашей?б)Млжно ли с уверенностью утверждать,что найдется хотыя бы 8 пеналов с одинаковым количеством карандашей?

Answer 1

А) точно, 7 пеналов будет

Answer 2

А) НАВЕРНО НО 20: 3 =6.(6)7

Answer 3

Analysis:

We are given that in each of the 20 pencil cases, there are either 4, 5, or 6 pencils. We need to determine if there are at least 7 pencil cases with the same number of pencils, and if we can confidently say that there are at least 8 pencil cases with the same number of pencils.

To solve this problem, we can use 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 pencil cases, and the pigeonholes are the possible numbers of pencils (4, 5, and 6).

Let's analyze each question separately.

a) Are there at least 7 pencil cases with the same number of pencils?

To determine if there are at least 7 pencil cases with the same number of pencils, we need to consider the worst-case scenario. In this case, the worst-case scenario is that there are 6 pencil cases with 4 pencils, 6 pencil cases with 5 pencils, and 6 pencil cases with 6 pencils. This totals to 18 pencil cases.

In the worst-case scenario, we still have 2 pencil cases remaining. Since we have only 3 possible numbers of pencils, we can conclude that there must be at least one number of pencils that appears in at least 7 pencil cases.

Therefore, it is true that there are at least 7 pencil cases with the same number of pencils.

b) Can we confidently say that there are at least 8 pencil cases with the same number of pencils?

To determine if there are at least 8 pencil cases with the same number of pencils, we again consider the worst-case scenario. In this case, the worst-case scenario is that there are 6 pencil cases with 4 pencils, 6 pencil cases with 5 pencils, and 6 pencil cases with 6 pencils. This totals to 18 pencil cases.

In the worst-case scenario, we still have 2 pencil cases remaining. However, since we have only 3 possible numbers of pencils, it is not possible to distribute these 2 pencil cases in a way that results in at least 8 pencil cases with the same number of pencils.

Therefore, we cannot confidently say that there are at least 8 pencil cases with the same number of pencils.

To summarize: a) It is true that there are at least 7 pencil cases with the same number of pencils. b) We cannot confidently say that there are at least 8 pencil cases with the same number of pencils.

Please let me know if anything is unclear or if you have any further questions!