100 Days streak problem

If 100 identical dices are rolled then the_number of different outcomes of the numbers appearing on the dice are?


The answer is 96560646.

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1 solution

Let a k a_{k} represent the number of dice that show the number k . k. Then we need to find the number of non-negative integer solutions to the equation

a 1 + a 2 + a 3 + a 4 + a 5 + a 6 = 100 a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} = 100 where 0 a k 100 0 \le a_{k} \le 100 for integers 1 k 6. 1 \le k \le 6.

This is a now a stars and bars problem, which according to Theorem two in this link has the solution

( 100 + 6 1 100 ) = ( 105 100 ) = 96560646 . \dbinom{100 + 6 - 1}{100} = \dbinom{105}{100} = \boxed{96560646}.

@Tanishq Varshney Nice problem, and congrats on the streak. :)

Brian Charlesworth - 6 years, 1 month ago

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Thank you sir ¨ \ddot \smile

Tanishq Varshney - 6 years, 1 month ago

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Congrats! @Tanishq Varshney Keep the streak alive!

Nihar Mahajan - 6 years, 1 month ago

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@Nihar Mahajan Congrats! I once got a 10,100 day streak. Oh, by the way, that's in binary (20). In binary, you'd have a 1,100,100 - a 1.1 million day streak!

vishnu c - 6 years, 1 month ago

Is there a theorem that can be used if the a i a_i aren't distinct? I know that for this problem, if the dice were distinct then the number of ways would be 6 100 6^{100} . But if you're trying to find the number of non-distinct ordered pairs ) a 1 , a 2 , a 3 , a 4 ) )a_1,a_2,a_3,a_4) such that a 1 + a 2 + a 3 + a 4 = 100 a_1+a_2+a_3+a_4=100 . What would you do then?

Trevor Arashiro - 6 years, 1 month ago

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If I understand correctly, I think we would be looking at the partitions of 100 100 of size 4. 4. Looking at a simpler version, say a 1 + a 2 + a 3 + a 4 = 5 a_{1} + a_{2} + a_{3} + a_{4} = 5 where the a i a_{i} 's are non-negative and non-distinct, we would have solution quadruplets

( 5 , 0 , 0 , 0 ) , ( 4 , 1 , 0 , 0 ) , ( 3 , 2 , 0 , 0 ) , ( 3 , 1 , 1 , 0 ) , ( 2 , 2 , 1 , 0 ) , ( 2 , 1 , 1 , 1 ) . (5,0,0,0), (4,1,0,0), (3,2,0,0), (3,1,1,0), (2,2,1,0), (2,1,1,1).

We could label this as P ( n ; k ) P(n;k) , the number of partitions of n n into at most k k parts. In this case we have P ( 5 ; 4 ) = 6. P(5;4) = 6. Here is the OEIS listing; there are some formulas but they're pretty scary, and the listing only goes up to 55. 55. However, that's the general idea. :)

Edit: Here is a good paper that I think addresses the matter at hand.

Brian Charlesworth - 6 years, 1 month ago

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Ok, wow, I thought there might have been something simple like stars and bars but guess not lol.

That's quite the interesting paper. It was a lot of fun to read through and I learned A LOT. Thanks for showing it to me :)

Trevor Arashiro - 6 years, 1 month ago

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@Trevor Arashiro You're welcome. That is a really good paper; I've bookmarked it for future reference. Partitions show up everywhere, (just like ϕ \phi ); I just learned about Bell numbers last week and have started to realize how significant they are. The more you know ..... :)

Brian Charlesworth - 6 years, 1 month ago

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