algebra q

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Here's a very longwinded interpretation/background. It's overkill - intentionally - as explanations go but hopefully allows you to now solve it yourself.

To use the pigeonhole principle here you need to work out two things, which are closely related:

What are the pigeonholes and how many are there?

You're trying to show that if you have 7 of something then there must be at least two satisfying a given criterion.

Just thinking here on how to go about the question - if all 7 fitted into their own pigeonholes the criteria is unlikely to be met.

So, we're looking for 6 pigeonholes.

This would force two to be in the same one.

Now we want "two in the same pigeonhole" to mean "two satisfying the criterion"

The criterion being those two are at most 1 cm apart. I.e. one lies within a certain distance of the other. If we fix one point, then the other lies within a certain area.

So, we're looking to divide the hexagon up into 6 areas, which will form the pigeonholes.

I think you should be able to have a go from there.

OK if we split the hexagon into 6 identical equalatiral triangles (1cm all sides) which will represent the pigeonholes calling thus k. the points in the hexagon are the pigeons calling this n. so n=7 and k =6 and 7/6 =2(taking upper bound). therefore 2 points will have to be within 1cm from each other since there are 6 equalatiral triangles and 7>6.

is the above enough to be an answer?

Posted from TSR Mobile

I would say:

We can divide the hexagon into six equilateral triangles. Since there are seven points, then at least one triangle must contain at least two points. The maximum distance any two points are apart in an equilateral triangle is the length of any side, viz 1cm. Hence there must be two points who are no more than 1 cm apart.

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Can see how you'd use pigeonholes here. If questions 2,4 on your handout are on pigeonholes, then it may be worth pursuing - some pigeonhole questions can be quite difficult to see.


Otherwise shubhro has the right idea, although his numbers are not all correct. Have a go.

OK this is what I done and then I'm pretty sure its wrong. I couldn't see the pigeon stuff so I just did it shubros way

Posted from TSR Mobile

This is really an S1 question on combinatorics, which you've already covered.

To start:

Consider how many possibilities for the first black ball?

Now once you've placed the first black ball, how many possibilities are there for the second one?

Etc.

PS: It's easier if you put your working on the forum, rather than an attached image, otherwise I can't quote it.

Neither ghostwalker or I think that it is a pigeonhole question. It's (fairly) elementary combinatorics.