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excluding christmas day.
explain how gordon can use a different route every day
explain how gordon can use a different route every day
Not sure what you mean. The simple answer is: because there are 364 different ways of navigating between grid locations (0,0) and (11,3).explain how gordon can use a different route every day
If you disagree, I would be interested to know how many different ways you believe there are between those two locations.
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You just need that eureka moment to see the logic of how the number of paths increases as you work down and across the grid.
How many routes are there from one corner of a single block to the diagonally opposite corner?
And for two blocks?
And four?
...
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Pascal's triangle - note the bordering '1's, if counting to a position then the border '1' is noted as zero.
364 at 3,14
Note the relationship between rows of numbers in the 'table' below, having cell border 'counters' changed to '1's similar to the triangle above.
I reckon the pupils will have completed the 'pattern' of values, using a calculator for addition only.
Lower right corner of 1st cell has two alternative journeys assosciated moving only East or South.... from there onward 'tis addition.
The formulae previously shown hold good
Add number at arrow base to number at it's head ... the sum is the new number to the right of arrow base....
--
364 at 3,14
Note the relationship between rows of numbers in the 'table' below, having cell border 'counters' changed to '1's similar to the triangle above.
I reckon the pupils will have completed the 'pattern' of values, using a calculator for addition only.
Lower right corner of 1st cell has two alternative journeys assosciated moving only East or South.... from there onward 'tis addition.
The formulae previously shown hold good
Add number at arrow base to number at it's head ... the sum is the new number to the right of arrow base....
--
- Joined
- 24 Sep 2005
- Messages
- 6,345
- Reaction score
- 268
- Country
I cannot imagine a child of 12 being expected to find this answer by calculation....
Therefore surely we must be into patterns - hence excel, I guess a sheet of paper would suffice- as Bolo has more or less shown ...
If each grid intersection is numbered by a particular method - think Pascal, then it should be possible for each number to reflect the sum of routes to that spot from the start point... I bet we'd find just one instance of the figure 364, (maybe not at 3,14 ) ... this will be the 'home' point... I think it will be the only point with the value 364 due to the wording of the Q .... "...but that on January 1st 2010, he would need to repeat a route already used..." I see that as confirmation that 364 is the max number of routes he could take. ( IMO Questions for children should not include 'red herrings' I do not believe this question does. Every part has a meaning. )
Why did I ignore, in my patterned spreadsheet a swathe of cells? Because I know that 3 from 14 has the same result as 11 from 14 ... 11 and 14 are not available together... If you see what I mean... However flawed !
Congratulations, empip, exquisite solution and diagrams.
One minor point (and sorry i did only scan the thread):
the answer is 4,12 not 3,11. You need to count the first row of ones as part of the solution. The row of ones represent the outer road that gordon can walk down. Next row represents Y*1, next row represents Triangular number series, next row represents Tetrahedral number series - which is where the 364 will occur.
This website explains all about pascals triangle and binomial coefficients.
They even have the example at the bottom of the page showing that :
t(12) = 364. (for tetrahedral numbers; i.e. the 4th collumn).
http://www.mathlesstraveled.com/?p=56
I remembered doing this at school about 15 years ago. We also did another variation about policeman and how many you need at the intersections to be able to see all the roads.
To be fair, to answer the question you only need to be rational about the permutations that can occur and perform some addition t produce the grid (pascals triangle). Being able to calculate any value in this grid is way beyond GCSE!!!!
the answer is 4,12 not 3,11. You need to count the first row of ones as part of the solution. The row of ones represent the outer road that gordon can walk down. Next row represents Y*1, next row represents Triangular number series, next row represents Tetrahedral number series - which is where the 364 will occur.
This website explains all about pascals triangle and binomial coefficients.
They even have the example at the bottom of the page showing that :
t(12) = 364. (for tetrahedral numbers; i.e. the 4th collumn).
http://www.mathlesstraveled.com/?p=56
I remembered doing this at school about 15 years ago. We also did another variation about policeman and how many you need at the intersections to be able to see all the roads.
To be fair, to answer the question you only need to be rational about the permutations that can occur and perform some addition t produce the grid (pascals triangle). Being able to calculate any value in this grid is way beyond GCSE!!!!
OK, i see. They want a set of coords from the station. The station being 0,0. then:
0 to 1 to 2 to 3
But from a maths point of view, the magic happens in the 4th column.
You have to walk 4 horizontal roads and 12 verticals.
Good fun!
0 to 1 to 2 to 3
But from a maths point of view, the magic happens in the 4th column.
You have to walk 4 horizontal roads and 12 verticals.
Good fun!
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