Arrangements and permutations

Question

Help please

ccooley · Answer

First, be careful with the language of the problem. Is that really the exact wording of the second part of the problem? A probability is the 'odds' that something will happen - so, you could ask this question and it would make sense:

2) If you randomly choose 5 letters in any order, what is the probability that they will include A, E, and I?

But if you ask this question:

2) What is the probability of choosing 5 letters in any order, if the five letters must include A, E and I?

It doesn't really make sense. You could say that the probability of choosing 5 letters is 0, because you don't want to choose any letters, so you won't! Or you could say that the probability of choosing 5 letters is 100%, because you're going to look at the letters ahead of time and choose A, E, and I, plus two other random letters.

You're probably interested, though, in the odds that A, E, and I will come up if you choose letters at random. So, you want to ask "what is the probability that the letters will include A, E, and I."

If this is a problem that someone gave you, I'd be wary of it - it definitely has confusing language.

But let's solve the edited version from above:

2) If you randomly choose 5 letters in any order, what is the probability that they will include A, E, and I?

To answer a probability question, start by writing this down on your paper:

probability = good cases / total cases

The total cases are the total number of ways you can choose 5 letters. The good cases are the cases where those letters include A, E, and I.

So, you really have two questions to answer:

How many ways can you choose 5 letters? That's 9c5, or 9!/(5!4!)

How many ways can you choose 5 letters including A, E, and I? Well, you're already stuck with A, E, and I, so really the only thing that can change is the remaining two letters. You want to count all of these possibilities:

A, E, I, M, C
A, E, I, M, H
A, E, I, M, N... etc.

You get one possibility for each set of the last two letters. How many different ways can those last two letters change? You're drawing them all from the pool M, C, H, N, S, T. That's six letters. You want two of them, in any order. So that's 6c2, or 6!/(2!4!).

Going back to the problem ...

We now have:

probability = good cases / total cases = 6c2 / 9c5

Simplify:

6c2 / 9c5 = (6!/(2!4!))/(9!/(4!5!)) = 6!4!5! / 9!2!4! = 6!5! / 9!2! = 5*4*3 / 9*8*7 = 5 / 3*2*7 = 5/42.

push12345 · Answer

) How many ways are there of arranging 5 letters from this word, if the order does matter?

9C5*5! Or 9P5

2) What is the probability of choosing 5 letters in any order, if the five letters must include A, E and I?

3 letters are already 
So one needs to select 2 from rest 6

So 6C2

Total=9C5

Probability=6C2/9C5

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Arrangements and permutations

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