Problems similar to the actual GMAT

[prep2max]

11c4*4! : 11c3*3! = 8 : 1

[arjmen]

1. 11p4/11p3 = (11!/7!)*(8!/11!) = 8!/7! = 8

2. (6-2)/(9-2) = 4/7

[akshayxyz]

Originally Posted by securitygeek

1. There is a set of characters, A B C D E F G H I J K. There are 4 character and 3 character codes to be made out of these characters. What is the ratio of the total number possibilities of the 4 character codes to the total possibilities of the 3 character codes?

Sorry for waking up the dead, but i believe the answer to the first question should be 11:1.. and not 8:1 for the following reasons
a) # of 4 character permutations:= 11*11*11*11 ( AAAA is acceptable code, nowhere it mentions that characters cant be repeated)
similarly # of 3 character codes = 11^3
so the ratio is 11:1

b) another approach, once again considering that duplicate characters are allowed..
# of 4 character codes = (# of choices for the 4th character) * (# of 3 character codes)..
we have 11 choices so.. its again 11:1

though I agree that if duplicates are not allowed, the ratio has to be 8:1

Hope I am not wrong. I am taking the test in 2 days , that said, any magic potions for Verbal

[blueskies]

Originally Posted by prep2max

1. 11c4*41 : 11c3*3!

2. 4/7

Why are you multiplying the combinations by 4! and 3! respectively?

I believe the solution is simply 11C4 / 11C3.

Ashayxyz, I see your point, but the feeling I get from the question is that letters are not to be duplicated. Of course, if "8:1" and "11:1" both show up in the answer choices... we're screwed.

[gonags]

Originally Posted by akshayxyz

Sorry for waking up the dead, but i believe the answer to the first question should be 11:1.. and not 8:1 for the following reasons
a) # of 4 character permutations:= 11*11*11*11 ( AAAA is acceptable code, nowhere it mentions that characters cant be repeated)
similarly # of 3 character codes = 11^3
so the ratio is 11:1

b) another approach, once again considering that duplicate characters are allowed..
# of 4 character codes = (# of choices for the 4th character) * (# of 3 character codes)..
we have 11 choices so.. its again 11:1

though I agree that if duplicates are not allowed, the ratio has to be 8:1

Hope I am not wrong. I am taking the test in 2 days , that said, any magic potions for Verbal

--------------
Securitygeek, what do you think about Akshayxyz's post? I think he's got a valid point.

Cheers!

[CTG1983]

I also think akshay is right.
Example:
How many telephone connections can be made with 5 digits from the natural numbers 1 to 9 inclusive?
The answer. is 9^5.

[mav]

Yeah, I agree that the proportion should be 11^4 : 11^3 = 11

At first, I think it's 11P4 : 11P3 = 8, but I think this approach is wrong.

[preity]

I think ETS will not give 11 in one of the answer choices..As the problem is not that clear about repetetions part. (Afterall its a problem solving Question. So go with the answers)

If they give 11 and 8 as answer choices then it will raise a conflict..I think they either give 8 or 11 as answer choices(not both)

What do u say securitygeek?????

[CTG1983]

Preity, Why do u think so?

[deren]

Originally Posted by akshayxyz

Sorry for waking up the dead, but i believe the answer to the first question should be 11:1.. and not 8:1 for the following reasons
a) # of 4 character permutations:= 11*11*11*11 ( AAAA is acceptable code, nowhere it mentions that characters cant be repeated)
similarly # of 3 character codes = 11^3
so the ratio is 11:1

b) another approach, once again considering that duplicate characters are allowed..
# of 4 character codes = (# of choices for the 4th character) * (# of 3 character codes)..
we have 11 choices so.. its again 11:1

though I agree that if duplicates are not allowed, the ratio has to be 8:1

Hope I am not wrong. I am taking the test in 2 days , that said, any magic potions for Verbal

hey akshayxyz;

first of all your solution is wrong because when you do this in your way ; 11*11*11*11 => solution set includes 4 AAAA and 4 BBBB and so on. Understand. So if we solve the problem in your way we need to substract 11 X 3 from our solution set. and the answer becomes something like that:

(11^4-33)\(11^3-22) .... and the solution is not even a integer...



I think this is a perm question and the real answer must be 8:1 ...




we don't know about the choices but probably your solution is