What is the difference between Selections and Arrangements/Permutaions?

22upon7 · #1 $Old$ November 5th, 2009, 09:47 PM

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

Robb · #2 $Old$ November 5th, 2009, 11:10 PM

Quote:

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 $5P3=60$ 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 $4P2=12$ ways that can happen. So $Pr(\mbox{H is first})=\frac{1\cdot 12}{60}=\frac{1}{5}$

Quote:

b. the letter H is chosen

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

Quote:

c. both vowels are chosen

So there is $2P2=2$ 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 $P(\mbox{both vowels are chosen})=\frac{2\cdot 9}{60}=\frac{3}{10}$

22upon7 · #3 $Old$ November 5th, 2009, 11:49 PM

Quote:

Originally Posted by Robb $View Post$

So, three letters are chosen randomly and arranged in a row. There are $5P3=60$ 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 $4P2=12$ ways that can happen. So $Pr(\mbox{H is first})=\frac{1\cdot 12}{60}=\frac{1}{5}$

Thanks Robb, I didn't understand this bit though, could you help me out with it?:

Quote:

Originally Posted by Robb $View Post$

Since you want the first letter to be a H, there is only one way that can happen, and there is $4P2=12$ ways that can happen.

Thanks again,

Dru

Robb · #4 $Old$ November 6th, 2009, 12:02 AM

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

22upon7 · #5 $Old$ November 6th, 2009, 12:28 AM

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

Thanks again,

Dru

Robb · #6 $Old$ November 6th, 2009, 12:48 AM

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

22upon7 · #7 $Old$ November 6th, 2009, 01:17 AM

No worries, Thanks again mate!