Counting with restrictions in die throws

Question

So I have this homework question.

A die is thrown 10 times, where the die has 6 faces labeled 1, 2, …, 6. Each outcome will be a sequence of 10 faces (i.e., 10 digits), where each face is one of the 6 values.

What is the number of possible outcomes where face 2 comes up exactly 3 times?

My first approach is to take the number of combinations of die that aren't 2 which would be $6^7$ and multiply it by the number of combinations of 10 die where 3 of them have 2 facing up: $3^{10}$. For a total combinations of $ 6^7 * 3^{10}$. Does this approach make sense or am I missing something?

Answer 1

A paraphrasing of the multiplication principle

To count the number of outcomes of a scenario, if you can describe every outcome uniquely via a sequence of steps such that the number of options available for a step does not depend on the specific choices made at previous steps (though the available options themselves may vary), the total number of outcomes is the product of the number of options available at each step.

Here, we use multiplication principle to count.

Step 1: Pick which three out of the ten available positions are occupied by twos (simultaneously).

Step 2a-2g: From left to right, in the unoccupied spaces pick one of the five numbers to occupy that space.

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For example, one such sequence of choices will result in the outcome $2~2~5~2~1~3~3~1~1~1$. To do this, in step one I may pick to have the twos occupy the first second and fourth spaces: $2~2~\underline{~}~2~\underline{~}~\underline{~}~\underline{~}~\underline{~}~\underline{~}~\underline{~}$

Then in step 2a: I pick the number $5$, leaving me with $2~2~5~2~\underline{~}~\underline{~}~\underline{~}~\underline{~}~\underline{~}~\underline{~}$

Then in step 2b: I pick the number $1$, leaving me with $2~2~5~2~1~\underline{~}~\underline{~}~\underline{~}~\underline{~}~\underline{~}$, and picking $3$ for 2c, $3$ for 2d, etc...

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Things to check:

• Is in fact every sequence of ten dice rolls with exactly three $2$'s achievable via at least one sequence of choices using the above setup?

• Is in fact every sequence of ten dice rolls with exactly three $2$'s achievable via at most one sequence of choices using the above setup?

• Is the number of options available at each step independent of the specific choices made earlier?

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Now, finally that we are hopefully on the same page, we can make it to the calculations.

How many ways can we complete step 1:? In how many ways can we pick three of the ten available positions to be occupied by the twos? We have ten available and we want to choose the three positions.

This can be done in $10$ choose $3$, i.e. $\binom{10}{3}=\frac{10!}{3!7!}$ ways. Not $10^3$ and not $\frac{10!}{7!}$. Remember that we are choosing them simultaneously, so we use the binomial coefficient. If we were to have broken this down as "pick the location of the first two" followed by "pick the location of the second two" etc... we would have overcounted because had we picked the locations in a different order it would have resulted in the same final sequence despite having used different choices at the steps to get there.

How many ways can you complete step 2a? How many ways can you pick a number from $\{1,2,3,4,5,6\}$ that isn't a $2$? I.e. How many ways can you pick a number from $\{1,3,4,5,6\}$?

$5$ ways.

And then 2b? 2c?... 2g? Taking all of these together as one step, how many ways then can you complete step $2$? I.e. How many ways can you fill the unoccupied spaces from left to right with a non-two number?

$\underbrace{5\cdot 5\cdots 5}_{7~\text{times}} = 5^7$ ways

So, our final total number of outcomes?

$\binom{10}{3}5^7$

A general formula for exactly $k$ twos rolled in $n$ rolls with an $s$-sided die (where $2$ is one of the available sides and all sides different)?

$\binom{n}{k}(s-1)^{n-k}$