Discrete Math - Pigeonhole Principle Problem

Click For Summary
SUMMARY

The discussion centers on applying the Pigeonhole Principle to a problem involving meal combinations. The problem states that a menu consists of 4 main dishes, 9 side dishes, and 6 desserts, with small meals comprising one main dish and two different side dishes, while large meals include one main dish, two different side dishes, and one dessert. The correct calculation for the total number of unique meal combinations is 4 * C(9,2) + 4 * C(9,2) * 6. To ensure at least two patrons order the same meal, one additional patron must be served, leading to the conclusion that the total number of patrons required is 4 * C(9,2) + 4 * C(9,2) * 6 + 1.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations (C(n, k))
  • Familiarity with the Pigeonhole Principle
  • Basic knowledge of meal composition in combinatorial contexts
  • Ability to perform calculations involving factorials and combinations
NEXT STEPS
  • Study the Pigeonhole Principle in greater depth to understand its applications
  • Learn about combinatorial mathematics, focusing on combinations and permutations
  • Explore real-world applications of the Pigeonhole Principle in various fields
  • Practice solving similar combinatorial problems involving meal combinations
USEFUL FOR

Students studying discrete mathematics, educators teaching combinatorial concepts, and anyone interested in applying mathematical principles to problem-solving scenarios.

Rockoz
Messages
30
Reaction score
0

Homework Statement


Suppose that a menu consists of 4 main dishes, 9 choices of side dishes, and 6 desserts. A small meal consists of one main dish and two different side dishes and no dessert. A large meal consists of one main dish, two different side dishes and dessert. How many patrons must be served to guarantee that two have ordered exactly the same meal?

Homework Equations


Pigeonhole Principle:
The pigeonhole principle states that if n items are put into m pigeonholes with n > m, then at least one pigeonhole must contain more than one item.

The Attempt at a Solution


I want to apply the pigeonhole principle here but I'm a bit unsure about how to break this problem down. The following is my attempt:

So I'm thinking I want to pick the minimal number of small meals and large meals, and then add one so I get just enough to have "more pigeons than pigeonholes" as the principle wants. So for just picking small meals, we have 4 large meals to choose from, and then C(9,2) possible combinations of side dishes and no desserts. So we have 4 * C(9,2) small meals. Next we want to pick the large meals and so we have 4 * C(9,2) * 6 possible large meal combinations. (4 large meals, C(9,2) side dish combinations, and 6 desserts). Next we add one so we can use the pigeonhole principle and state that two patrons must have ordered exactly the same meal.

So I think the answer is: 4*C(9,2) + 4*C(9,2)*6 + 1.

Again, thank you for your time. Please let me know if my thinking is incorrect and why.
 
Physics news on Phys.org
Hi Rockoz
I believe you are correct, never heard of this 'principle' before though :)
One thing: why are you talking about 'minimal number of small and large meals' ?
this is the total number of different meals, if anything, it is maximal (maximum number of non repeating meals)

Cheers..
 
oli4 said:
Hi Rockoz
I believe you are correct, never heard of this 'principle' before though :)
One thing: why are you talking about 'minimal number of small and large meals' ?
this is the total number of different meals, if anything, it is maximal (maximum number of non repeating meals)

Cheers..

Right, that makes sense that it should be called maximal. That's what I was trying to describe. Bad word choice by me.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
21K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
3
Views
4K