# Thread: What is the difference between Selections and Arrangements/Permutaions?

1. ## Probability Application help

Hi,

I'm revising for my Maths exam on Tuesday and I really need help with this question:

Three letters are chosen at random from the word "HEART" and arranged in a row. Find the probability that:

a. the letter H is first
b. the letter H is chosen
c. both vowels are chosen

I think this is how you work a. out:

HEART has 5 letters, so 1/5 = 0.2

I'm not sure how to work b. and c. out though, I think it needs permutation and/or combinations but I can't work out the correct answer.

Any help would be greatly appreciated, and I'm sure I'll have more questions to come.

Thanks,

Dru

2. Three letters are chosen at random from the word "HEART" and arranged in a row. Find the probability that:

a. the letter H is first

So, three letters are chosen randomly and arranged in a row. There are $\displaystyle 5P3=60$ ways they can be arranged... Since you want the first letter to be a H, there is only one way that can happen, and there is $\displaystyle 4P2=12$ ways that can happen. So $\displaystyle Pr(\mbox{H is first})=\frac{1\cdot 12}{60}=\frac{1}{5}$

b. the letter H is chosen
Letter H can now be in 1 of 3 places, so
$\displaystyle P(\mbox{Letter H is chosen})=\frac{3\cdot 12}{60}=\frac{3}{5}$

c. both vowels are chosen
So there is $\displaystyle 2P2=2$ ways to arrange the 2 vowels, and the remaining 3 characters can be arranged in any of the 3 positions, so there are 9 ways for the other letter to be positioned.
So $\displaystyle P(\mbox{both vowels are chosen})=\frac{2\cdot 9}{60}=\frac{3}{10}$

3. Originally Posted by Robb
So, three letters are chosen randomly and arranged in a row. There are $\displaystyle 5P3=60$ ways they can be arranged... Since you want the first letter to be a H, there is only one way that can happen, and there is $\displaystyle 4P2=12$ ways that can happen. So $\displaystyle Pr(\mbox{H is first})=\frac{1\cdot 12}{60}=\frac{1}{5}$
Thanks Robb, I didn't understand this bit though, could you help me out with it?:

Originally Posted by Robb
Since you want the first letter to be a H, there is only one way that can happen, and there is $\displaystyle 4P2=12$ ways that can happen.
Thanks again,

Dru

4. Just the permuation formula,
$\displaystyle \frac{4!}{(4-2)!}=\frac{24}{2}=12$

5. yes, but how did you get those values? The 4 and the 2?

Thanks again,

Dru

6. Oh, sorry. Orignally you had 5 letters to choose from, and you were selecting 3. But because the first one is a H, you are now selecting 2 letters from the remaining 4.
So the probability is just the ways you can arrange H with 2 other letters, divided by the total number of ways you could arrange 3 letters from 5

7. No worries, Thanks again mate!