Permutations and Combinations

Question

A fair $6$-sided die is rolled 5 times and the result is recorded for each roll. How many different results are possible? Of the possible results, in how many ways can there be a result containing exactly $2$ rolls of a $4$?

For the first part I got $6^5 = 7776$ different results are possible.

For the second part I am lost at what steps I should take. I was thinking $(\frac16)^2$ for rolling a $4$ twice, but the result doesn't make sense to me of $\frac1{36}$??

Any guidance would be greatly appreciated.

Answer 1

You know, from the first part, that there are $7776$ total possibilities. How many of these have exactly two fours?

Imagine the rolls ordered from $1$ to $5$. We can choose which rolls the fours will be rolled on as a combination, since there are five possible slots and two fours to place:

$$\binom{5}{2} = \frac{5 \cdot 4}{2}= 10$$

After we've chosen the fours, we have three slots remaining and five possibilities for each slot (since there can't be any more fours). Our final answer is $$10 \cdot 5^3 = \boxed{1250}$$

Answer 2

Think of your 5 rolls as a string of numbers e.g. 64323, 53614, 25364, etc. In the first case the numbers can be anything (use # as wildcard), so we can write it as #####. The answer is 6*6*6*6*6 = 6^5 as you have said, since for each of the 5 rolls there are 6 possible outcomes.

In the case where 2 rolls are required to be 4, the string will be of the form 44###, #4#4#, ###44, etc.. Since order doesn't matter, examine 44###. There is 1 possible way of rolling a 4 on a 6-sided die, so the answer is 1*1*6*6*6 = 6^3.

Answer 3

The two $4$s can occur in any of $\binom52=10$ positions out of the five rolls. For each choice of where to place the $4$s, we have $5^3$ choices for what the other three rolls are. That gives us a total count of $10\cdot5^3=1250$.

When you're talking about using $\frac16$ for rolling each $4$, you're thinking of calculating a probability. This is just a counting problem. Many times, the easiest way to calculate a probability is to do one counting problem for the denominator, and another for the numerator. In this case, by doing these two counting problems we've discovered that the probability of rolling exactly two $4$s is $\frac{1250}{7776}$.