Please help with these problems

[gre320hex]

Originally Posted by skawal

according to your reasoning should the ans be A?

Sorry, I mixed up my previous post. This is how it should have read.. Actually, I read a better way to answer this kind of a question elsewhere. Notice, that every time a greater term is getting added than the term that is being subtracted. And, we see that the first 2 terms are 1-1/2= 1/2. From, after that we have 1/2 + ( 1/2-1/4+1/5..) ie. 1/2 + ( bigger term - smaller term ). Therefore, S is greater than 1/2

[skawal]

kudos gre320hex,
i think, to solve problem like this, one need a keen observation.

[gre320hex]

Originally Posted by skawal

kudos gre320hex,
i think, to solve problem like this, one need a keen observation.

Ok, I need to admit this was not my idea. I had to recollect what I had read elsewhere

[gre320hex]

Originally Posted by aimgre08

yes that is correct for that combination problem and

for the other combination problem on dollars,(am not sure if i am correct!!)
Total no of ways of giving 7 dollars to 3 people and each should have min.1 dollar is similar to no. of ways of giving 4 dollars to 3 people = 3^4

Total no. of ways of giving 7 dollsrs to 3 people and 1 getting min 3dollars, other two a minimum of 1 dollar is similar to no of ways of giving 2 dollars to 3 people = 3^2

so the value in Column B is greater.

I think you are right in the first part. No.of ways with everyone having atleast 1 dolloar = (3C1)^4 But, total no of ways with atleast one person having 3 dollars should be as follows: No of ways 3 dollars can be given to a any of the 3 people = 3 ways. Then, no of way remaining 4 dollars can be given to 3 people is (3C1)^4 . Since, since, for every dollar you can choose among one of the 3. Therefore, total would be 3*(3C1)^4. Therefore, A is greater. ( I am wondering why my paragraphs are getting reformatted, and getting squished into one big blob of text. Anyone having similar problems )

[gre320hex]

Originally Posted by skawal

for the combination problem,i dont know whether my reasoning is correct or not.however,below my reasoning.

if 2 integers is selected from 10 integers there should be 45 possible selections.

the question asked you to select the possible selections that consists of one odd and one even integer.

selection consists of only even integers=5c2=10
selection consists of only odd integers=5c2=10

so possible selections that consists of one odd and one even integer
=45-(10+10)=25

Another easy way would be just saying
number of ways to choose 1 digit to be even 5C1
number of ways to choose 1 digit to be odd 5C1

Therefore, total = 5C1*5C1 = 25
I think both ways probably takes the same time. Just wanted to point a different way to solve the problem.

[skawal]

Originally Posted by gre320hex

Another easy way would be just saying
number of ways to choose 1 digit to be even 5C1
number of ways to choose 1 digit to be odd 5C1

Therefore, total = 5C1*5C1 = 25
I think both ways probably takes the same time. Just wanted to point a different way to solve the problem.

I like your approach.

[Goldust]

About the dollar problem, I think it's easiest to solve using the pigeonhole principle.

Let's say we have to find a way such that no person has 3 dollars. We give each person 2 dollars. But, uh-oh, what do we do with the remaining dollar?

The critical point is that there is NO way in which each person receives at least one dollar and no person receives 3 dollars.

Hence, the two columns are equal.

[gre320hex]

Originally Posted by Goldust

About the dollar problem, I think it's easiest to solve using the pigeonhole principle.

Let's say we have to find a way such that no person has 3 dollars. We give each person 2 dollars. But, uh-oh, what do we do with the remaining dollar?

The critical point is that there is NO way in which each person receives at least one dollar and no person receives 3 dollars.

Hence, the two columns are equal.

Goldust, I am not sure I got your answer. The questions asks "The number of ways" and not if it is possible. The answer is B, and cannot be C. Do we have an OA on this?


Edit:
Ok, I see where I made a mistake. I overlooked the initial condition posed by the question - " every person atleast had one dollar"
In that case
ways to distribute remaining 4 dollars = 3^4.
ways to distribute 3 dollars to one person, and remaing 3 dollars = 3.3^3.
Therefore, C.
Goldust, you have a good point. It was hard for me to see the answer from that perspective. Thanks a lot!

[skawal]

the OA is C.no doubt.
but i have confusion about the problem.

[gre320hex]

Originally Posted by skawal

the OA is C.no doubt.
but i have confusion about the problem.

skawal, whatz you confusion?