December 3rd, 2017 at 5:04:15 PM
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I have 7 dice
each die has face values of {1,1,1,5,10,25}
They are my coin dice and I roll them.
How many ways for each sequence below?
How to easily figure it.
1) 1,1,1,1,1,1,1
2) 5,1,1,1,1,1,1
3) 5,5,1,1,1,1,1
4) 10,5,1,1,1,1,1 (I counted 10206 ways for this sequence)
I can't do it
I have to count the ways
thank you
Sally
each die has face values of {1,1,1,5,10,25}
They are my coin dice and I roll them.
How many ways for each sequence below?
How to easily figure it.
1) 1,1,1,1,1,1,1
2) 5,1,1,1,1,1,1
3) 5,5,1,1,1,1,1
4) 10,5,1,1,1,1,1 (I counted 10206 ways for this sequence)
I can't do it
I have to count the ways
thank you
Sally
I Heart Vi Hart
December 3rd, 2017 at 6:53:49 PM
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For the last one, 3^5 x 7 x 6 = 10,206.
December 3rd, 2017 at 7:13:39 PM
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Quote: mustangsally
How to easily figure it.
1) 1,1,1,1,1,1,1
2) 5,1,1,1,1,1,1
3) 5,5,1,1,1,1,1
4) 10,5,1,1,1,1,1 (I counted 10206 ways for this sequence)
If I am understanding the problem correctly:
First, if you have N numbers, and N1 of them are one number, N2 are a second number, ..., and Nk are a kth number, then the number of ways they can appear is N! / (N1! x N2! x ... x Nk!).
For #1, there are 7 1s, so they can appear once; on each die, 1 appears 3 times, so the total is 1 x 37 = 2187.
For #2, there are 7! / (1! 6!) = 7 ways for the numbers to appear; the one 5 appears once on that die, and each of the six 1s appear 3 times on each of those dice, so the total is 7 x 1 x 36 = 5103 ways.
For #3, there are 7! / (2! 5!) = 21 ways for the numbers to appear; each of the two 5s appears once on each of those dice, and each of the five 1s appears three times on each, so the total is 21 x 12 x 35 = 5103 ways.
For #4, there are 7! / (1! 1! 5!) = 42 ways for the numbers to appear; the 10 appears once on that die, the 5 appears once on that die, and each of the four 1s appears three times on each, so the total is 42 x 1 x 1 x 35 = 10,206.
Here's one for you: 25, 10, 10, 5, 5, 1, 1
There are 7! / (1! 2! 2! 2!) = 630 ways for the numbers to appear; the 25 appears once on its die, each of the two 10s appears once on each of those dice, each of the two 5s appears once on each of those dice, and each of the two 1s appears three times on each of those dice, so the total is 630 x 1 x 12 x 12 x 32 = 5670 ways.
December 3rd, 2017 at 10:23:00 PM
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#2 and #3 got me.Quote: ThatDonGuyFor #2, there are 7! / (1! 6!) = 7 ways for the numbers to appear; the one 5 appears once on that die, and each of the six 1s appear 3 times on each of those dice, so the total is 7 x 1 x 36 = 5103 ways.
For #3, there are 7! / (2! 5!) = 21 ways for the numbers to appear; each of the two 5s appears once on each of those dice, and each of the five 1s appears three times on each, so the total is 21 x 12 x 35 = 5103 ways.
same # of ways and I stopped.
I left out one important step after that.
thank you so much
easy when one sees how
Sally
I Heart Vi Hart