Two integers x and y are chosen without replacement out of the set {1,

Let us take different scenarios:

1) x=1: y can take 9 values.....
\(1^y\)..... y can be anything from 2 to 10

2) y=1: x can take 8 values.......
\(x^1\)....x can take any value from 2 to 9, as 10^1=10, a 2-digit number.

3) x=2, possibilities \(2^1, \ \ 2^3\). But y=3 is the only possible value, as 2^1 is already considered above in (1) and rest values of y will make x^y a 2-digit number..

4) x=3, possibilities \(3^1, \ \ 3^2\). But y=2 is the only possible value, as 3^1 is already considered above in (1) and rest values of y will make x^y a 2-digit number....

Total: 9+8+1+1=19 values

Possible values without any restriction = 10*9=90

So, prob=\(\frac{19}{90}\)

D
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I agree an disagree with this answer...

What I disagree with is that, since it is without replacement you cannot say that there are still 8 values when y=1.

That because if you take x=1, then you get:
1 to the powers of 2,3,4,5,6,7,8,9,10 (which is equal to 1, one digit, so accpeted). 10 integers used (including 1).
You can use 2 to the power of 3, but then not 3 to the power of 2 (same numbers used). So, 10+1=11.

If y=1, then you get:
2,3,4,5,6,7,8,9 to the power of 1 (which is equal to 2,3,4,5,6,7,8,9, accepted). However, 1^2 and 2^1 use the same 2 numbers (1 and 2) that you used above for x=1. So, since there is no replacement, these numbers shouldn't be available. Right?

On the other hand, I understand your approach and that if you have y=1 then you can only use 8 numbers.
I also agree with the use of 2 and 3. However, to me, the rule of replacement has been violated...

So, I would say that A should be the right answer...

It would be great if this question could be discussed and we could have the correct answer...

hi ..
what replacement would have done to the ans is that 1^1 and 2^2 would have also been part of the answer..
also, 1^2 and 2^1 are different and correct as the value of x and y is changing in both the answers...
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What I understand though from the stem is that x and y are both integers that are part of the specified set. So, shouldn't they be there only once?

Then, if you choose one of these integers for x you immediately cannot choose the same y (as you said already), but also, if you have chosen let's say 2 for x, then 2 is taken and is not available to be chosen for y.

Let me ask, do you know that the correct answer is indeed D? Then I will accept what you are saying as a fact, otherwise, I am just not sure..

then what do you think total probabilities are as per your understanding of question...
how does the no reach 90...
taking x as 2 and y as 1 finishes one probability and we start fresh when we take y as 1 and x as 2..
this is exactly what without replacement means...
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Hey,

This is a good question. The number would reach 90 because you have 10 possible numbers and 9 of them are single digits. So, 9*10=90.

Also, I am not saying that I am right here. I am trying to understand the question and how "without replacement" would work in this case.

If you look at it as 2 different possibilities, then you are right, you can start over with y. In this case, you would have 19 possibilities.

My question here is, does "without replacement" mean that once one set of numbers (one x and one y) is used it cannot be used again? Or is it that we should find the possible sets of 2 numbers starting with x first and continuing with y?

And my guess is that this was also the purpose of these 2 options..

Set 1 = {1^2,1^3,...,1^10} = 9
Set 2 = {2^1,3^1,4^1,5^1,6^1,7^1,8^1,9^1} = 8
Set 3 = {2^3} = 1
Set 4 = {3^2} = 1

19 / 90

If x = 1, y could = 2, 3, 4, 5, 6, 7, 8, 9, 10
If x = 2, y could = 1, 3
If x = 3, y could = 1, 2
If x = 4, y could = 1
If x = 5, y could = 1
If x = 6, y could = 1
If x = 7, y could = 1
If x = 8, y could = 1
If x = 9, y could = 1
x cannot = 10 as x^9 is single-digit

Number of possible values is therefore 19. Total possibilities (10*9) is 90. So answer = D.

VERITAS PREP OFFICIAL SOLUTION:

Since there will not be replacement when picking this set, there are 10 first numbers to be selected and the remaining 9 numbers to be selected next, for a total of 10 * 9 = 90 possible outcomes.

Again, noting that there cannot be repeat numbers, you'll need to omit the chances of 1 to the 1st power and 2-squared as single-digit results. With that in mind, the possibilities are:



For a total of 19/90.
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ok, I tried to count all the outcomes

x can be 1, then y can be 2,3,4,5,6,7,8,9 = 1*8=8 ways
x can be 2, then y can be 1, 2 or 3 = 3 ways.
x can be 3, then y can be 1 or 2 = 2 ways.
x can be 4, then y can be 1 only - 1 way
x can be 5, then y can be 1 only - 1 way
x can be 6, then y can be 1 only - 1 way
x can be 7, then y can be 1 only - 1 way
x can be 8, then y can be 1 only - 1 way
x can be 9, then y can be 1 only - 1 way

19 ways.
since we can arrange all numbers in 90 ways, then the probability will be 19/90.

tough one, took me >3m to solve.

Hi mvictor,

To me also It took more than 3 minutes to solve to get the right answer. the moment i looked at the question i again thought of bringing a logic in mind rather than counting. after wasting 15 sec, i had to count. I am sure GMAT does give such kind of questions.

Do you think there is any logical approach to such questions.?

Prob = (# fav outcomes in which (X)^y = one digit) / (Total Possible Outcomes)

DEN: We have 10 numbers from which we can choose 2 without replacement

“10 choose 2” = 10! / (2! 8!). = 45 different ways we can choose 2 numbers to place in X and Y

And

Then for each of these 45 different combinations, we can arrange the 2 numbers chosen in 2! = 2 ways between the X and Y Variable

DEN = 90

At this point, I found it easiest to just count the cases in which (X)^y will result in a single Digit Value

(1st)
When X Base = 1

Y can be any of the other 9 values

9 possibilities


(2nd)
When Y Power = 1

The X Base can take any of the other Integers EXCEPT 10 ——-> everything up to (9)^1 will result in a one Digit value


8 more favorable cases


(3rd)
(2)^3 = 8
(3)^2 = 9

2 more cases

(9) + (8) + (2) = 19

Answer

19/90

Trap: watch out that you don’t chose any cases such as (2)^2 or (1)^1 ——-> this is selection without replacement

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