Simple combinatorics dice problem

Question

How would you explain, how ${5\choose4}$ corresponds to the pattern $6-6-6-6-3$ when throwing five dice. The pattern is meant to form a sum of $27$ in total.

What I do understand is that you could shift the place of $3$ to form $4$ different patterns than the one above. However that would mean that this combination (${5\choose4}$) somehow acts as a permutation in some sense. But then I would like to know what do I choose from (5), and why do I choose that what I choose (4).

Answer 1

The expression

$${5 \choose 4}$$

counts the number of ways that you can roll $5$ dice to come up with $4$ sixes and a single three.

You have five dice. That's the $5$. You're choosing $4$ of them to have a value of six. The remaining one has a value of three.

Note that you can also count them as

$${5 \choose 1}.$$

You're choosing $1$ die of the $5$ to have a value of three, and the others have a value of six.

In general,

$${n \choose k} = {n \choose n-k}.$$

Answer 2

You have gone dice, four of them are sixes and one is a three. ${5 \choose 4} = 5$ is the number of ways to pick which of the dice show $6$.

Answer 3

In this particular instance, imagine that you are allowed to sit all five dice in front of you, and then fix the values of each. It is also important to assume that the dice are in some way distinct, (different size, different colors, etc) otherwise we do not have any natural orderings.

Now, we know that we would like to obtain the sum 27 out of four 6s and one 3, and we want to show that there are (5 choose 4) ways to do this. The most natural way to see this is to simply choose 4 out of the 5 dice, and fix their values to be 6. There are exactly (5 choose 4) ways to do this. The final die must then be the 3 by default.

Answer 4

Yes, binomial coefficients are not only combinations, it have many meanings depending what you are doing.

You can understand easily if you think the binomial coefficient as a multinomial coefficient that is used to count permutations with repetition but just for 2 sets/groups, where you can think repetition as a class or set of something (not just ONE number... it can means a group or set of something).

$$\binom{n}{k_1,k_2,...,k_n}=\frac{n!}{k_1!k_2!\cdots k_n!};\ where\space\ n=k_1+k_2+\cdots+k_n$$

In your case 5 is the total number of elements, and 4 and 1 are the multiplicities for each group, the group of 6 and the group of 3.

$$\binom{5}{4}=\frac{5!}{1!4!}$$

Additional example: for this sequence a-a-a-a-b-b-c-c-c-d the amount of permutations is

$$\binom{10}{4,2,3,1}$$