In maths, what is your odds chance of winning this competition
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In maths, what is your odds chance of winning this competition

[From: ] [author: ] [Date: 13-01-30] [Hit: ]
.. x 17) when we cared about order. Then we would reduce this by a factor of (9 x 8 x .........

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It depends if you are concerned with order or not. When you say "combinations", that implies that order doesn't matter eg as with the lottery, but perhaps that is not the case since your example looks to be expressed with a specific order.

If order is important it is simply 26^9, since each letter has 26 outcomes and we choose letters 9 times.
26^9 = 5.4 x 10^12

If order isn't important the number falls, but it isn't so easy to calculate because you allowed repeats.

Without repeats our initial answer would have been (26 x 25 x ... x 17) when we cared about order. Then we would reduce this by a factor of (9 x 8 x ... x 1) because each set of numbers can be arranged in that many unique ways (but they all count the same). This gives about 53 x 10^6 ie considerably smaller.

However, with repeats it gets harder because for example,

abcdefghi can be put into (9 x 8 x ... x 1) different looking arrangements
aaaaaaaab can only be put into 9 different looking arrangements
aaaaaaaaa can only be put into 1 arrangement

Thus the number of repeats, and the number of different repeats, in each outcome, all has to be accounted for. Doing that is frankly a mess and beyond most people's interest. Well, mine anyway.

Maybe just say, more than 53 x 10^6 and less than 5.4 x 10^12 :-)

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