If 100 identical dices are rolled then the_number of different outcomes of the numbers appearing on the dice are?
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@Tanishq Varshney Nice problem, and congrats on the streak. :)
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Thank you sir ⌣ ¨
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Congrats! @Tanishq Varshney Keep the streak alive!
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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!
Is there a theorem that can be used if the a i aren't distinct? I know that for this problem, if the dice were distinct then the number of ways would be 6 1 0 0 . But if you're trying to find the number of non-distinct ordered pairs ) a 1 , a 2 , a 3 , a 4 ) such that a 1 + a 2 + a 3 + a 4 = 1 0 0 . What would you do then?
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If I understand correctly, I think we would be looking at the partitions of 1 0 0 of size 4 . Looking at a simpler version, say a 1 + a 2 + a 3 + a 4 = 5 where the 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 ) .
We could label this as P ( n ; k ) , the number of partitions of n into at most k parts. In this case we have 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 5 5 . However, that's the general idea. :)
Edit: Here is a good paper that I think addresses the matter at hand.
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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 :)
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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 ); I just learned about Bell numbers last week and have started to realize how significant they are. The more you know ..... :)
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Let a k represent the number of dice that show the number 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 = 1 0 0 where 0 ≤ a k ≤ 1 0 0 for integers 1 ≤ k ≤ 6 .
This is a now a stars and bars problem, which according to Theorem two in this link has the solution
( 1 0 0 1 0 0 + 6 − 1 ) = ( 1 0 0 1 0 5 ) = 9 6 5 6 0 6 4 6 .