Good permutation question. (1 Viewer)

[CHUDYMASTER]

This was in the 4 U test at my school, but it's more harder 3 U work.

How many ways can you arrange 6 letters taken from the word AUSTRALIA?

I got a fair idea of what the answer is, but I'd like to see what the rest of you think.

 

[spice girl]

Originally posted by CHUDYMASTER
This was in the 4 U test at my school, but it's more harder 3 U work.

How many ways can you arrange 6 letters taken from the word AUSTRALIA?

I got a fair idea of what the answer is, but I'd like to see what the rest of you think.

7! + 6*5*4*3*(6C2) + 6*5*4*(6C3)
= 12840

 

[McLake]

I agree with spice girl.

7! if 'a' is picked 0 or 1 times.

6*5*4*3*(6C2) if 'a' is picked 2 times.

6*5*4*(6C3) if 'a' is picked 3 times.

so 7! + 6*5*4*3*(6C2) + 6*5*4*(6C3) = 12840

 

[CHUDYMASTER]

Hmm, I did not get that. Don't you add 7! twice for both 0 and 1 a's scenarios?

I worked it out a totally different way and did not quite get what you got.

 

[spice girl]

Originally posted by CHUDYMASTER
Hmm, I did not get that. Don't you add 7! twice for both 0 and 1 a's scenarios?

no, for both 0 and 1 'A's, we use the same scenario: the 7P6 selection, as it takes into account whether you choose the 'A' or not.

 

[CHUDYMASTER]

I think I follow now. Thanks!

 

[Saintly Devil]

Originally posted by McLake
I agree with spice girl.

7! if 'a' is picked 0 or 1 times.

6*5*4*3*(6C2) if 'a' is picked 2 times.

6*5*4*(6C3) if 'a' is picked 3 times.

so 7! + 6*5*4*3*(6C2) + 6*5*4*(6C3) = 12840

I don't get it. Why is it 6C2 and 6C3? I thought it would be (8P6 /2) and 9P6/3.

so i thought it would be 7P6+ 8P6/2! + 9P6/3! = 20,167 ways

 

[McLake]

Originally posted by Saintly Devil

I don't get it. Why is it 6C2 and 6C3? I thought it would be (8P6 /2) and 9P6/3.

Since the "a's" are eliminated there are only 6 to choose from. Its C instead of P because order is an unordered selection (since the order comes during the multiplication, if that makes any sense ...)

 

[Saintly Devil]

here's another............

What is the probability that any one out of 3 people will be in the same group if they are to be randomly placed in one of four groups?

this includes if all three people are in the same group as well as if only two are in the same group.

 

[spice girl]

Re: here's another............

Originally posted by Saintly Devil
What is the probability that any one out of 3 people will be in the same group if they are to be randomly placed in one of four groups?

this includes if all three people are in the same group as well as if only two are in the same group.

ridiculously ambiguous. you can interpret this question in many different ways.

Assuming the people are called A, B, C

"any one of 3 people", does that mean any of A, B, C? or does it mean any particular person (e.g. just A)?

"be in the same group" as who? As any other person of the three?

I'll assume that it's any particular person (A only), and being in the same group as B and/or C.

i.e., we're checking for the possibility of the existence of the groups AB, AC, ABC.

Place A in a group first (doesn't matter which, the problem is symmetric)

Then the P(AB OR ABC) = 1/4, since the chance of placing B in the same group as A is 1/4

Same reasoning why P(AC OR ABC) = 1/4

But now you've double-counted P(ABC), so deduct one of this. P(ABC) = 1/16 since P(B in same group as A) = 1/4, and P(C in same group as A) = 1/4.

So, P(AB, AC, ABC) = 1/4 + 1/4 - 1/16 = 7/16

 

[Saintly Devil]

hmm......that's strange...........

because i just listed out all the possibilities (yeah, sorry about the ambiguity) and i got 4/5.............20 possible arrangements in total, and out of those only 5 where none of them are together,

so doesn't that make the probability of any one of them being together (no, i meant the probability of ANY of them being together, not just person A with person B and C) 4/5?

 

[spice girl]

You mean you want P(AB, BC, AC, ABC) ?

Hmm...this is 1 - P(A,B,C separate)

P(A,B,C separate) = 4*3*2/4*4*4 = 6/16 = 3/8

So P(AB, BC, AC, ABC) = 1-3/8 = 5/8

I don't know how you got 20 arrangements, there should be 64 (4*4*4), 4 different options to place each of A,B,C.