MHB Count Arrangements of 16 Balls in 4 Drawers-At Least 3 Each

  • Thread starter Thread starter CStudent
  • Start date Start date
  • Tags Tags
    Count
CStudent
Messages
14
Reaction score
0
Hey.

* How many ways we can insert 16 similar balls to 4 drawers such that in every drawer we have at least 3 balls?

So I know we have to insert at first 3 balls to every drawer and the remainder is 4 balls.
But how can I calculate the amount of possibilities to insert 4 balls to 4 drawers without any limit?

Thanks.
 
Physics news on Phys.org
The number of ways $k$ indistinguishable balls can be put into $n$ drawers is equal to the number of multisets of cardinality, or size, $k$ consisting of $n$ distinct elements. (I explain why below.) This number is often denoted by $$\left(\!\!\!\binom{n}{k}\!\!\!\right)$$ and is called a multiset coefficient. Just like the binomial coefficient $$\binom{n}{k}$$ equals the number of subsets, or combinations in combinatorics parlance, of size $k$ taken from a set of size $n$, the multiset coefficient $$\left(\!\!\!\binom{n}{k}\!\!\!\right)$$ equals the number of multisets, or combinations with repetitions, of size $k$ whose elements come from a set of size $n$. Note that unlike the number of combinations, in the latter case $k$ can be greater than $n$ because elements in a multiset can be repeated.

There is a one-to-one correspondence between multisets of size $k$ drawn from a set of size $n$ and $n$ drawers that have $k$ balls combined inside them. For example, if $n=4$ and $k=5$, the multiset $\{1,1,3,4,4\}$ corresponds to the situation where drawer 1 has two balls, drawer 3 has one ball and drawer 4 has two balls.

Finally, $$\left(\!\!\!\binom{n}{k}\!\!\!\right)=\binom{n+k-1}{k}$$. This is explained in Wikipedia and in the linked article there.
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.
Back
Top