## Interesting Maths Stuff (using wolframalpha) We have 8 balls: white, black, orange, red, blue, green, yellow and purple = {1,2,3,4,5,6,7,8} A) Here I am drawing a ball from a bag, placing it on the ground, recording its color and then placing the ball back in a bag. Shake the bag to mix up the balls. I then repeat the process three more times to get a total of four colors recorded down on my paper.

I then go out to a storehouse full of colored balls and get the same colored balls I have recorded down on the paper. How I place the balls depends on if the order is important or not–and also this results in a different range of possibilities (numbers of combinations or permutations) as shown below.

Just out of interest, our number system works in the same way–we have 10 symbols. For example if we wrote down 4 random symbols (numbers), the number could range from 0000 to 9999 or we have 10^4 (10,000) possible sets of four symbols (numbers).

### random Number = 2314 for example

But if the order of the placement of the numbers was not important, then this number “1234” for example would be the same as these numbers:

• …., 2134, 2314, 2341
• 1324, 3124, 3214, 3241
• 1432, 4132, 4312, 4321
• 1243, 2143, 2413, 2431
• 1342, 3142, 3412, 3421
• 1423, 4123, 4213, 4231

Just out of interest, when a number like 0000 occurs, it only has one possible arrangement. So with numbers that have a the same number occuring more then once, the numbers of possible re-arrangements will be less than sets of four numbers that do not repeat. The same principle will also apply for the color balls we draw out of a bag (when replacement of drawn ball is allowed).

Thus the possible sets of four numbers is greatly reduced to 330 possible permutations. In the above example, you could achieve that effect by writing down four random numbers and then reordering ordering them from the lowest number to the highest number.

In the examples below (as mentioned above) we draw a ball out of a bag record it color and then place it back in the bag (shake the bag to mix the balls up) and then draw another ball out of the bag. We continue this procedure until we have recorded the colors of four balls. 1) Combination: If I could use the same colored ball more than once, and order is important: then how many different groups of 4 balls would there be: n^4 … 8^4 = 4,096

reference: 2) Permutation: If order is not important, and I could use the same colored ball more than once: then how many groups of four could there be (n+r-1)!/(r!*(n-1)! … (8+4-1)!/(4!*(8-1)!=  330

Reference:

B) 1) Combination: If I drew the balls from a bag randomly without replacement and the order is important: n!/(n-r)! … 8!/(8-4)! = 1,680

reference: 2) Permutation: If I drew balls out of a bag without replacement, and the order is not important: n!/(r!*(n-r)!) … 8!/(4!*(8-4))!=  70

Reference:

With wolfram Alpha you can try out different numbers to see what happens in the above equations.

Here is a variation of this problem: If we have a car number-plate with three numbers and three letters, what is the maximum number of number-plates we could ever have:

10*10*10*26*26*26 = 10^3 * 26^3 = 17,576,000  Click here for the wolfram alpha calculation . In today’s world it is quite possible that 18 million vehicles might exist in a country, so the solution might be to make number-plates by using more numbers and/or letters-for example a number-plate with 4 numbers and 4 letters would give a maximum number of 4,569,760,000 number-plates. Commands put into wolframalpha: integrate 25*sin(x^2*2) integrate 25*Sin(x^2*2)

Commands put into wolframalpha: polyhedron regular dodecahedron Dodecahedron

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