GPrep 2

[RonPurewal]

parthian7, yes, that method works. but, you appear to be thinking that it's somehow fundamentally different from the "brute force" method. it's not.
in fact, your method is exactly identical to the "brute force" method, except in that your method is more visual in nature (i.e., you made a tree rather than just a list of possibilities).

Also notice that this way there's no need to write down all possibilities, which would be brutal if we had more than 4 letters. We only draw it out for 1 case and multiply by 4, since they are identical.

you can do the same thing with the brute force method. (this should be clear, because, after all, your method actually is the brute force method.)

* list out all possible cases in which letter "a" goes into the correct envelope but the others don't:
acdb
adbc
... that's it

* then, just as you did, multiply by 4 for reasons of symmetry.

[parthian7]

parthian7, yes, that method works. but, you appear to be thinking that it's somehow fundamentally different from the "brute force" method. it's not.
in fact, your method is exactly identical to the "brute force" method, except in that your method is more visual in nature (i.e., you made a tree rather than just a list of possibilities).

Also notice that this way there's no need to write down all possibilities, which would be brutal if we had more than 4 letters. We only draw it out for 1 case and multiply by 4, since they are identical.

you can do the same thing with the brute force method. (this should be clear, because, after all, your method actually is the brute force method.)

* list out all possible cases in which letter "a" goes into the correct envelope but the others don't:
acdb
adbc
... that's it

* then, just as you did, multiply by 4 for reasons of symmetry.

well to be honest, I only use the combination/permutation method for these problems, which I think is different than brute force (I only used the tree diagram above to help people understand the reasoning, which probably made it look like the brute force..)

otherwise, I would just go:

solving for A:
I have 4 slots available and above each slot I write down the number of possibilities:

1 * 2 *1* 1 = 2
_ | _ | _ | _

for all letters: 2 *4 = 8

answer: 8/24 = 1/3

------------------------------------------------------------

or easier yet:
P = # of desired possibilities / total # of possibilities
(due to symmetry) = # of desired possibilities (for A) / total number of possibilities (for A)
= 1*2*1*1 / 3! = 1/3

[tim]

i can't actually tell whether your approach works. can you be more specific about what you did with the 1*2*1*1?

[parthian7]

i can't actually tell whether your approach works. can you be more specific about what you did with the 1*2*1*1?

sorry what do you mean by "what you did with 1*2*1*1?"

If you mean how I arrived at it? like I explained in the previous post:

I have 4 (envelopes) spots and 4 letters. For when A goes into Ea and B not into Eb, C not into Ec and D not into Ed, I would have the following # of possibilities for every spot:

1 ----- 2 -----1-----1-----
Ea-----Eb-----Ec-----Ed

Hence total number of desired possibilities for this case=1.2.1.1=2

If you mean what I did with the result? well it was the number of desired output, so I plugged it into the numerator: in the first approach I multiplied it by 4 before doing so and in the 2nd approach I plugged it in directly.

Hope it's clearer now? :)

[RonPurewal]

i can't actually tell whether your approach works. can you be more specific about what you did with the 1*2*1*1?

sorry what do you mean by "what you did with 1*2*1*1?"

If you mean how I arrived at it? like I explained in the previous post:

I have 4 (envelopes) spots and 4 letters. For when A goes into Ea and B not into Eb, C not into Ec and D not into Ed, I would have the following # of possibilities for every spot:

1 ----- 2 -----1-----1-----
Ea-----Eb-----Ec-----Ed

Hence total number of desired possibilities for this case=1.2.1.1=2

If you mean what I did with the result? well it was the number of desired output, so I plugged it into the numerator: in the first approach I multiplied it by 4 before doing so and in the 2nd approach I plugged it in directly.

Hope it's clearer now? :)

basically, the verdict here is this:
yes, your 1x2x1x1 works.
BUT,
it doesn't work like a simple product of probabilities, because of the restrictions on the possibilities here.
in other words, your math works, but you actually need the tree diagram (= the brute force method) to show why it works in this case.

[simplyankush]

Quickly

Probability: Number of favourable events
________________________
Number of possible events

In this case, number of ways the letters can be arranged is 4!
Let the letters be called A , B , C , and D
Now consider the scenario where only one letter can be in its correct position. This letter can be chosen in 4 ways.

The next letter has 3 positions to choose from. But one is its correct position. So it can be placed in 2 positions.

The next two letters can be arranged in only 1 way such that they are in incorrect positions.

So this makes it

Probability = 4 * 2*1
---------
4!

which is 1/3

[tim]

thanks!