this is the person who asked the question.
there are 3 questions:
a) Mr Wong bought some presents for his pupils in his class. He bought some pens, erasers and files. Every pupil is allowed to choose 2 presents. What is the least number of pupils to ensure that two of them will choose the same type of presents?
Assume that the first pupil takes a pen and an eraser, the second pupil takes a pen and a file, and the third pupil takes a file and an eraser, than all the ways of choosing the presents will be up, so when added one more pupil which is 4 pupils altogether, we can ensure that two of them will choose the same type of presents.
b) A box contains 10 blue ice-cream sticks, 20 red ice-cream sticks, 8 yellow ice-cream sticks, 15 green ice-cream sticks and 25 purple ice-cream sticks. How many must I choose to ensure that 12 ice-cream sticks are of the same colour?
When we choose all 10 blue ice-cream sticks, 11 red, all 8 yellow, 11 green and 11 purple, we do not have 12 of anything yet, but if we choose any one remaining, we will have 12 of that colour. Thus, 10+11+8+11+11+1=52 will guarantee at least 12 ice-cream sticks are of the same colour.
c) Show that if we select 51 distinct numbers from 1 to 100, at least two are consecutive numbers. (Hint: You can prove by contradiction)
Because in 100 numbers we can select 50 all odd or 50 all even and there will be no consecutive pair, picking another number must be of the opposite parity and so it will be consecutive with one of your first 50 numbers chosen.
and i have to state the generalistions from the solutions above. please help!
The main idea is the most important one! If you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then there must result at least one pigeon hole with more than one pigeon. Expand the idea perhaps using a finite example. 21 pigeons and 10 holes. Try and see how at least one hole must have more than 2 pigeons. Then think about generalizing it. The real power comes from seeing this all in action. A commonly used example is: Among any N positive integers, there exists 2 whose difference is divisible by N-1.
The simple form of it states: If m pigeons are put into m pigeonholes, there is an empty hole iff there's a hole with more than one pigeon. In more formal math language it says: Let |A| denote the number of elements in a finite set A ( also known as its cardinality). For two finite sets A and B, there exists a 1-1 correspondence f: A -->B if and only if |A| = |B|.
Let's look at a simple example first such as 3 consecutive integers. We want to show that it is divisible by 3.Take 4,5, and 6 and of course 6 is divisible by 3.The reason for this is can be seen using the pigeonhole principle.When an integer is divided by 3, possible remainders are 0, 1, and 2. It follows that everyinteger can be expressed in one of the forms 3k, 3k + 1, and 3k + 2 where k is an integer.So if we have any three consecutive integers, one of them must be divisible by 3.Let's look at how the pigeonhole principle applies. Suppose we have 3 consecutive integers are non are divisible by 3. Think of the pigeon holes as 3k, 3k + 1, and 3k + 2, now this means no numbers are in the 3k hole and two of them must be in either the 3k+1 hole or the 3k+2 hole. But this contradicts that they are consecutive integers.So for any n, let the pigeon holes be nk, nk+1,... nk+(n-1) and these exhaust the multiples of n. Now if you take n consecutive numbers, you must have a least 1 number in the nk pigeon hole or else they are not consecutive.
Counting Principle is one of them
Type your answer here... I AM A DWARF AND I AM DIGGING A HOLE! DIGGING DIGGING HOLE! DIGGING DIGGING HOLE! HOLE HOLE, DIGGING DIGGING HOLE! DIGGING DIGGING HOLE! DIGGING DIGGING HOLE! HOLE HOLE, DIGGING DIGGING HOLE!
A pigeon hole :)
The main idea is the most important one! If you have fewer pigeon holes than pigeons and you put every pigeon in a pigeon hole, then there must result at least one pigeon hole with more than one pigeon. Expand the idea perhaps using a finite example. 21 pigeons and 10 holes. Try and see how at least one hole must have more than 2 pigeons. Then think about generalizing it. The real power comes from seeing this all in action. A commonly used example is: Among any N positive integers, there exists 2 whose difference is divisible by N-1.
Pigeon-Hole. JT
a hole between the right and left ventricle
The Canadian band Pigeon Hole consisted of Dusty Melo and Marmalade. The band was started in 2010 in Canada. It is considered a Canadian Hip Hop band.
Garden Web is a great resource for people to find instructions on how to build a pigeon hole. They have several options and provide printable versions for people to have handy.
Learn to "pigeon hole" problems. Pigeon holes are small compartments in which pet pigeons can make nests. When you pigeon hole your problems, you keep them in the situation in which they occurred. For example, suppose you had a dreadful day at work. The machine you ran broke down and you didn't accomplish your goals for the day. You should do your best to keep this problem at work so it doesn't interfere with your friendships, home life, studies, and other activities.
well a$$ hole. a pigeon just ran into the window
O-S-S- - 1957 Operation Pigeon Hole 1-13 was released on: USA: 19 December 1957
The cast of Horse in the Hole - 2013 includes: Damien Devaney as Charlie Pigeon Kian Murphy as Boy
To put an object in a safe location where it can be found easily.
Many do. Such generalisations are never true.