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

One kind of proof using an argument by contradiction:

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

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

My question 1:

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

Another kind of proof using a direct argument:

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

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

Another question 2:

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

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

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

My question 1:

How did the author deduce that "Then $\displaystyle 8$ or more computers are used by $\displaystyle 2$ 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 $\displaystyle k$ be the number of computers used by $\displaystyle 3$ or more students.[We must show that $\displaystyle k \geq 5$] Because each computer is used by at most $\displaystyle 6$ students, these computers are used by at most $\displaystyle 6k$ students(by the contrapositive form of the generalized pigeonhole principle). The remaining $\displaystyle 12 - k$ computers are each used by at most $\displaystyle 2$ students.

Another question 2:

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