1. Good post? |
1) seven on-dollar bills are to be distributed among 3 people so taht each gets at least 1.
A: number of ways to distribute so that each gets at least \$3
b: total number of ways to distribute the bills

2) which is boht the mean and the meadian of 7 consecutive integers, the least of whcih is x?

x + 4
x + 3
X+ 2
x+28/7
x+3/2

2. Good post? |
Truman,

For number 1 a) there is no way to distribute 3 dollars to each of three people when you only have 7.. If this were to work you would need at least 9 dollars. (maybe you wrote the question in error)

For 1 b) is the answer 15? If so I will explain.

For number 2, the answer should be x+3. if you know that you have 7 consecutive integers, then you have an odd amount of numbers. So the middle or median will be the same as the mean. if x is the first of the consecutive integers and you add 3 to x you will be at the 4 term, or the middle term.

Example. lets say you have 1 2 3 4 5 6 7. If x is 1 then x +3 = 4 which is the middle, or median. But in this case it is also the mean.. To check if you want, add all the terms up and divide by 7. you should have 28/7 = 4.

3. Good post? |
Hi gmatfordays, truman,

For 1 a) 3C3 is the no of ways three people will get at least one bill while the remaining 4 can be distributed in 4C3 ways.

Am not sure, correct me if I'm wrong

Shivani

4. Good post? |
Hi Shivani.. I agree with 1 a) for sure.. However 1 b) I don't know.. I too came up with that, but actually felt it was not correct.. 4c3 is actually 4. But I think it is 4c3 x 3.. Why, because 4c3 give you 4 and there are four ways you can change it up. But that is just one set (not sure of my logic here). So multiplying it by 3 shows the 12 different ways you could distribute 4 extra dollars.

I did the brute force method, as I too was unsure about this 4c3, and I came up with 12. I just now figured out that it might be 4c3 x 3, which equals 15.

Hopefully someone else can validate.

5. Good post? |
Here is what I found on an old post.

Posted - 2003 Apr 29 : 04:45:45
--------------------------------------------------------------------------------

Hi Milan,

By brute force,

A B C
-----
1 1 5
1 2 4
1 5 1
1 4 2
1 3 3
2 4 1
2 1 4
2 2 3
2 3 2
3 1 3
3 3 1
3 2 2
4 1 2
4 2 1
5 1 1

therfore 15 ways
and all of them have atleast one having atleast \$3

http://www.TestMagic.com/forum/topic.asp?TOPIC_ID=4230

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