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If a red bag contains ball s numbered 1 to 20, and a blue bag contains ball s numbered 21 to 41.

 

Can you move a ball from the blue bag to the red bag, and then another ball from the red bag to the blue bag, and again from the blue bag to the red bag, and so on, in such a way as to make the contents of the red bag go through all the possible combinations without repetition?

 Mar 10, 2016
edited by MATHBITCH  Mar 10, 2016
 #1
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Beats me.

 

There are 20 b***s in the red bag and 21 b***s in the blue bag

 

 

 

 There is 41C20 combination are possible

 

269128937220 = 2.6912893722e11

 

That is a lot of combinations!

 Mar 10, 2016
 #4
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Actually it is probably more than that because sometimes the bag as 20 b***s and sometimes it has 21 b***s.

I was only looking at the combinations of 21 b***s ://

 

 

 

There are going to be a lot of (wipe out)  stars in this answer!     LOL

Melody  Mar 10, 2016
 #2
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Mind blown lol

 Mar 10, 2016
 #3
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Hi guest.  It is mind blowing!

 

Here is something else that might b**w you mind.Say your mum gives you 1c allowance for the first week of January.

That is pretty useless you say - might as well throw it away!

BUT

then each week after that she gives you double as much as she geve you last times.

So in week 2 you get 2c

week3         4c

week 4        8c

You have still only got a total of 1+2+4+8 = 15c

 

week 5       16c

week 6        32c

week 7       64c

week 8       $1.28

week 9       $2.56

 

by the time you get to the end of the year - week 52 you will get     2^51 approx    2.2518*10^15 cents

That is     week 52          $ 22,518,000,000,000

 

How is that for mind blowing :)

 Mar 10, 2016
 #5
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I agree that is mind blowing melody just not as mind blowing as I can imagine someone wuth that amount of money. Back to the ball question, imagine if you actually tried recording that many combinations to find one without repetition

 Mar 10, 2016
 #6
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Yes it would take a while LOL cheeky

 Mar 10, 2016

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