# Thread: Generalized and contrapositive form of Pigeonhole principle and a math problem

1. ## Generalized and contrapositive form of Pigeonhole principle and a math problem

A math problem:

There are $42$ students who are to share $12$ computers. Each student uses exactly $1$ computer and no computer is used by more than $6$ students. Show that at least $5$ computers are used by $3$ or more students.

One kind of proof using an argument by contradiction:

Suppose not. Suppose that $4$ or fewer computers are used by $3$ or more students.[A contradiction will be derived.]Then $8$ or more computers are used by $2$ or fewer students.

.........
(End of first proof)

My question 1:

How did the author deduce that "Then $8$ or more computers are used by $2$ or fewer students"?

Another kind of proof using a direct argument:

Let $k$ be the number of computers used by $3$ or more students.[We must show that $k \geq 5$] Because each computer is used by at most $6$ students, these computers are used by at most $6k$ students(by the contrapositive form of the generalized pigeonhole principle). The remaining $12 - k$ computers are each used by at most $2$ students.

.........
(End of this proof)

Another question 2:

Why's that "The remaining $12 - k$ computers are each used by at most $2$ students."?

2. ## Re: Generalized and contrapositive form of Pigeonhole principle and a math problem

Originally Posted by x3bnm
Suppose not. Suppose that $4$ or fewer computers are used by $3$ or more students.[A contradiction will be derived.]Then $8$ or more computers are used by $2$ or fewer students.

My question 1:

How did the author deduce that "Then $8$ or more computers are used by $2$ or fewer students"?
Let U be the set of all computers and let A be the set of all computers used by 3 or more students. The problem asks to prove that |A| (the cardinality of A) is >= 5. Suppose this is not the case, i.e., |A| <= 4. Then |U \ A| = |U| - |A| = 12 - |A| >= 8. Also, each computer in U \ A is used by at most two students by the definition of A.

Let $k$ be the number of computers used by $3$ or more students.[We must show that $k \geq 5$] Because each computer is used by at most $6$ students, these computers are used by at most $6k$ students(by the contrapositive form of the generalized pigeonhole principle). The remaining $12 - k$ computers are each used by at most $2$ students.

Another question 2:

Why's that "The remaining $12 - k$ computers are each used by at most $2$ students."?
Again, k is the number of all computers with at least 3 users; the other 12 - k computers have at most 2 users by the definition of k.

3. ## Re: Generalized and contrapositive form of Pigeonhole principle and a math problem

Originally Posted by emakarov
Let U be the set of all computers and let A be the set of all computers used by 3 or more students. The problem asks to prove that |A| (the cardinality of A) is >= 5. Suppose this is not the case, i.e., |A| <= 4. Then |U \ A| = |U| - |A| = 12 - |A| >= 8. Also, each computer in U \ A is used by at most two students by the definition of A.

Again, k is the number of all computers with at least 3 users; the other 12 - k computers have at most 2 users by the definition of k.
Thank you emakarov. That answers my question.