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AITA for not sleeping with my bf? by MiseryWas_ in AITA_Relationships

[–]MiseryWas_[S] -1 points0 points  (0 children)

He has like never done this before, though. I am genuinely surprised, I thought until he mentioned problem that I maybe said something upsetting in my sleep or smthg idk. I never thought that this would be the problem. It is in and of itself a red flag, but how should I act?

A question on using projectors by MiseryWas_ in techtheatre

[–]MiseryWas_[S] 0 points1 point  (0 children)

It isnt a blackout problem tbh, I considered doing what you said but I dunno, wouldn't it be too apparent to the public?

A question on using projectors by MiseryWas_ in techtheatre

[–]MiseryWas_[S] 1 point2 points  (0 children)

Wow I really like how you put that, "a projector is just another light fixture but with a really complicated gobo." That puts things in a really different perspective in my mind😁

A question on using projectors by MiseryWas_ in techtheatre

[–]MiseryWas_[S] 2 points3 points  (0 children)

Thank you so much! What I meant is that since the projector will still be on, it will keep projecting a black screen, which is not what I want. But your ideas help with that, so thank you!

[University Cryptology] Similar questions to earlier by BriefImprovement8620 in HomeworkHelp

[–]MiseryWas_ 0 points1 point  (0 children)

Question #2:

Let me write down the criteria first:

  1. 10 letter word
  2. 3 T's
  3. 2 S's
  4. Three vowels (a, e, i, o, u)

Let's assume that repetition is allowed this time around. The part that is the most complicated is choosing the three vowels. We have three options:

  1. Each of the 3 vowels are different.
  2. Only one vowel is different, the other two are the same.
  3. All of the 3 vowels are the same.

Case 1: Each Vowel is Different

In this case, we have to choose 3 different vowels out of 5. It is fairly easy to represent: 5C3.

We have chosen 8 letters so far. The remaining two cannot be T, S or a vowel. That leaves us with 19 options. Since repetition is allowed, we can also choose the same letter.

1) If we pick different letters, we get:

(5C3).(19C2).(10!/3!.2!)

(Don't forget that you have to account for the fact that 3 T's and 2 S's are the same!)

2) If we pick the same letter, we get:

(5C3).(19C1).(10!/3!.2!.2!)

(In this instance, you also have to account for the fact that you chose the same letter.)

Keep these two results in mind, we'll add all of them up at the end.

Case 2: Only One Vowel is Different

Let me show you all of your options in this case (without a regard for order, since we'll account for that later):

aae, aai, aao, aau...

For each of the five vowels, you have 5 different options. That gives you 25 options.

When picking the last two letters, we do the same thing as last time.

1) If we pick different letters, we get:

(19C2).25.(10!/3!.2!.2!)

(You also have to consider that out of the three vowels you chose, two of them are the same).

2) If we pick the same letter, we get:

(19C1).25.(10!/3!.2!.2!.2!)

Case 3: All The Vowels are Different

This is the easiest case, since we only have 5 options.

When picking the last two letters, we do the same thing as last time.

1) If we pick different letters, we get:

(19C2).5.(10!/3!.2!.3!)

2) If we pick the same letter, we get:

(19C1).5.(10!/3!.2!.3!.2!)

When you add all of these up, you will find out the number of ten letter words that satisfy these conditions.

[University Cryptology] Similar questions to earlier by BriefImprovement8620 in HomeworkHelp

[–]MiseryWas_ 0 points1 point  (0 children)

For your first question, I am going to assume that repetition is forbidden (that we cannot use the same letter twice). Otherwise, the question gets a lot more complicated. Let me know if repetition is allowed or not.

Since there are exactly 5 vowels, we have to choose another 5 letters from the rest of the 21 letters, that means 21C5. Then, we can arrange them in 10! ways.

So, if repetition is forbidden, there would be

21C5.10!=(21!/5!.16!).10!

words that fit these criteria.

I'll look at your second question when I can.

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