A single die is rolled. how many ways can you roll a number less than 3 then an even and then an odd

Assuming the die is just marked even and odd rather than with numbers, there are eight orderings. They range from all odd to all even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In your formula:

$$\frac{n!}{(n-r)!}$$

You are missing that you don't care about the orders of the duplicates. So you need

$$\frac{n!}{(n-r)!r!}$$

The $n$ is the total number of rolls, the $r$ is the number of either evens or odds (it's symmetric). But to get the total number of orderings, you need to add these:

$$\sum\limits_{r=0}^n\frac{n!}{(n-r)!r!}$$

Now substitute 3 for $n$.

$$\sum\limits_{r=0}^3\frac{3!}{(3-r)!r!}$$

Unrolling that, we get

$$\frac{3!}{3!0!} + \frac{3!}{2!1!} + \frac{3!}{1!2!} + \frac{3!}{0!3!}$$

or

$$1 + 3 + 3 + 1$$

So we have one all odds, three with one even, three with two evens, and one all even. That's eight total.

Another way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be.

Of those eight, how many fit your parameters? Exactly one, OOE. So one in eight or $\frac{1}{8}$.

As others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll. There's three rolls, so $(\frac{1}{2})^3 = \frac{1}{8}$.

If you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can. But it is much easier to think of them as just odd or even. Because you don't want to try write this out for $6^3 = 216$ orderings. And in the end, you will get the same basic result. You will have twenty-seven OOE orderings, which is again one eighth of the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $\frac{27}{216} = \frac{1}{8}$.

Permutations leads you down a harder path. It's easier to think just in terms of probability or even ordering.

pasagot pox (x + 5) = 21/x + 2x/5 = 2​

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what does the prolem ask you to find​

Find the 15th term of the A S 1 5 9 13​

Complete the statement using always, sometimes or never 1.) An even number of negative factors will _______________ produce a product that is negative … . 2.) An odd number of negative factors will _______________ produce a product that is negative.

Find the sum of the first 20 positive even numbers.​

Positive integers greater than 4

please help me to answer this questions​

1. How many students only like Volleyball.. 2. How many students like football com Wall 3. How many students like football or Volleyball 1. How many d … o not like to both football or Volleyball • How many students like Volleyball HOW many students do not like both football or How many students do not lolike Volleyball. How many students like both football and Volle How many students like football How many students do not like football.​

6.) ×2+6×-11=0 7.)×2+14×=32​

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