In how many ways will the sum of the number of 1’s and 2’s (total of occurrences of 1's and 2's) equ

Scott Martinez

Scott Martinez

Answered question

2022-06-02

In how many ways will the sum of the number of 1’s and 2’s (total of occurrences of 1's and 2's) equal the sum of the number of 3’s, 4’s, 5’s and 6’s, after n rolls, assuming n = 2i for some positive integer i? What is the probability of this occurring?

Answer & Explanation

Ashlynn Mcmillan

Ashlynn Mcmillan

Beginner2022-06-03Added 5 answers

n = 2 i, where i N, so we want i occurrences of 1's and 2's, and i occurrences of 3's, 4's, 5's and 6's.
For simplicity of notation, denote the number of occurrences of a number x by n x
Lets do the 1's and 2's first.
n 1 can be anything from 0 to i (giving i + 1 choices), but whatever the value of n1, it must be that n 2 = i n 1 i.e. we have i + 1 choices for the 1's and 2's.
Now lets look at the 3's, 4's, 5's and 6's.
Again, n 3 can be anything from 0 to i (giving i + 1 choices). But this time, we have more choices for n 4 as it can go from 0 to i n 3 . Similarly, once we have decided on n 4 , n 5 can only go from 0 to i n 3 n 4 . Finally, n 6 must be whatever is left, so n 6 = i n 3 n 4 n 5 . Written out mathematically, the number of choices here is
n 3 = 0 i n 4 = 0 i n 3 n 5 = 0 i n 3 n 4 1 = ( i + 3 ) ( i + 2 ) ( i + 1 ) 6
So combining them together, we get a total of
( i + 1 ) ( i + 3 ) ( i + 2 ) ( i + 1 ) 6

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