Arrangements: Frans was given 3 white and 3 black square tiles of the same size to...

danandmichelle

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Jun 20, 2018
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Hi there, I'm a mum trying to help my 9 and 11 year old practice some ICAS questions and having trouble with some of them. You are not supposed to use algebraic formulas to solve any of the questions.

The first question is:

Frans was given 3 white and 3 black square tiles of the same size to fill a rectangle divided into 6 of the same size square tiles. How many different arrangements can he make? I know the answer is 20 but without sitting there rearranging the 6 tiles (which isn't practical in a test situation anyway), I'm not sure how they got it!

Thank you for any help!
 

Subhotosh Khan

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Hi there, I'm a mum trying to help my 9 and 11 year old practice some ICAS questions and having trouble with some of them. You are not supposed to use algebraic formulas to solve any of the questions.

The first question is:

Frans was given 3 white and 3 black square tiles of the same size to fill a rectangle divided into 6 of the same size square tiles. How many different arrangements can he make? I know the answer is 20 but without sitting there rearranging the 6 tiles (which isn't practical in a test situation anyway), I'm not sure how they got it!

Thank you for any help!
If all the tiles were different colors (say black, white, red, blue, green and yellow). How many ways can you arrange tiles in that rectangle?
 

Dr.Peterson

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Nov 12, 2017
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Hi there, I'm a mum trying to help my 9 and 11 year old practice some ICAS questions and having trouble with some of them. You are not supposed to use algebraic formulas to solve any of the questions.

The first question is:

Frans was given 3 white and 3 black square tiles of the same size to fill a rectangle divided into 6 of the same size square tiles. How many different arrangements can he make? I know the answer is 20 but without sitting there rearranging the 6 tiles (which isn't practical in a test situation anyway), I'm not sure how they got it!

Thank you for any help!
I'm not sure where the boundary is between using "algebraic formulas" and not. I would solve this using combinations (it is 6C3, because you are choosing 3 of the 6 places to put a black tile), and that is normally done using something that could be called a formula: 6!/(3!3!) = (6*5*4)/(3*2*1) = 20. There are ways to count without that formula, but I'd consider them harder. The best way I can think of for someone who hasn't learned combinations is what is called an "orderly list": write out each arrangement in such a way as to be sure to cover all possibilities. If done well, that really is not too slow; but I probably wouldn't write a test that demanded that.

Then again, I'm surprised that this would be expected by any method at those ages. Do you know what they have been taught in this area, that they might be expected to use? Have they said anything about what thoughts the problem evokes?

I checked out what the ICAS is, and it looks like it may include an occasional challenge like this. Can you show us where you got this sample problem, and what level it is for? Where do they mention avoiding formulas? Any further information might help us see what might be expected.
 

Jomo

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Dec 30, 2014
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Hi there, I'm a mum trying to help my 9 and 11 year old practice some ICAS questions and having trouble with some of them. You are not supposed to use algebraic formulas to solve any of the questions.

The first question is:

Frans was given 3 white and 3 black square tiles of the same size to fill a rectangle divided into 6 of the same size square tiles. How many different arrangements can he make? I know the answer is 20 but without sitting there rearranging the 6 tiles (which isn't practical in a test situation anyway), I'm not sure how they got it!

Thank you for any help!
Imagine numbering the positions that the tiles can go in 1 thru 6. The number of arrangements will not matter if you put the tiles into the form of a rectangle or put them in a line (Please think about that). So we want to arrange 6 tiles in a line. If the tiles were all different, then their will be 6! ways of arranging the tiles. But we have three tiles that are black and three tiles that are white. An arrangement might be WWBWBB where white tiles are in the 1st, 2nd and 4th position and black tiles in the 3rd, 5th and 6th position. So the real question is how many ways can you arrange 6 letters if you have three w's and three b's. The answer is 6!/(3!*3!)= 20
 

danandmichelle

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Jun 20, 2018
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I'm not sure where the boundary is between using "algebraic formulas" and not. I would solve this using combinations (it is 6C3, because you are choosing 3 of the 6 places to put a black tile), and that is normally done using something that could be called a formula: 6!/(3!3!) = (6*5*4)/(3*2*1) = 20. There are ways to count without that formula, but I'd consider them harder. The best way I can think of for someone who hasn't learned combinations is what is called an "orderly list": write out each arrangement in such a way as to be sure to cover all possibilities. If done well, that really is not too slow; but I probably wouldn't write a test that demanded that.

Then again, I'm surprised that this would be expected by any method at those ages. Do you know what they have been taught in this area, that they might be expected to use? Have they said anything about what thoughts the problem evokes?

I checked out what the ICAS is, and it looks like it may include an occasional challenge like this. Can you show us where you got this sample problem, and what level it is for? Where do they mention avoiding formulas? Any further information might help us see what might be expected.

Thank you Dr Peterson, I have attached the actual practice test, please refer to question 38. This is an Australian Year 5 Maths ICAS test (10-11 year olds). The test starts off with questions aimed to 2 or 3 grades below and finishes with questions aimed to 2 or 3 grades above. I doubt permutations and combinations is taught even 2 or 3 grades above Year 5. Also, refering to the ICAS website, it states the following for Papers A-E (Year 5 are Paper C).


  • Calculators are not permitted.
  • Formal algebra is not tested. The emphasis is put on pattern, structure and puzzles.
  • Formal geometry is not tested (except for a few items in Paper E). The emphasis is put on spatial skills.
https://www.eaa.unsw.edu.au/icas/subjects/mathematics

So given that formulas would not be expected, I was hoping there would be another explanation of how to work it out (for example, say the question is "There are 6 children needing to be arranged into 3 seats, so the first seat can hold any of the 6, the second can hold any of the rest of the 5 etc...that would be a simple explanation without using formulas). I just don't understand myself how to explain this one as not all 6 objects are different - 3 are black and 3 are white. Also, if the only way is to sit and write out the different arrangements, do you have a tip on how to do it so it's not so confusing!
 

danandmichelle

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Jun 20, 2018
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Imagine numbering the positions that the tiles can go in 1 thru 6. The number of arrangements will not matter if you put the tiles into the form of a rectangle or put them in a line (Please think about that). So we want to arrange 6 tiles in a line. If the tiles were all different, then their will be 6! ways of arranging the tiles. But we have three tiles that are black and three tiles that are white. An arrangement might be WWBWBB where white tiles are in the 1st, 2nd and 4th position and black tiles in the 3rd, 5th and 6th position. So the real question is how many ways can you arrange 6 letters if you have three w's and three b's. The answer is 6!/(3!*3!)= 20
Thank you Jomo :smile:
 
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