Help with a comparation problem

[Goldust]

Col A says that at least one person receives at least $3. If this number was less than the total number of ways of distribution, then there would be some ways in which no person receives $3 or more, right? Try finding such a way. At the most, each person can have $2 then. But there is no way you can distribute $7 in this manner, without giving $3 to one person. Hence, col A and col B are equal. If you have studied statistics, you will recognize this as an application of the pigeonhole principle.

[mystic87]

Originally Posted by prithibi

seven-one dollar prblm
answer: B
m i wrong? someone explain plz

They are indeed the same. I went about this question the wrong way, it shouldn't require any type of calculation like Goldust pointed out:

Seven $1 bills and 3 ppl, you simply can't give all of them less than 3.
3-2-2
3-1-3
4-1-2
5-1-1
Those are the only possiblities (with different orders as well of course). You see how you HAVE TO include a number that's at least 3 or bigger? that's why A and B are in fact equal. Hope this helps.

[mystic87]

Question for Goldust:

Purely for statistics purposes, if we were to arrange these possibilities then when do we know to multiply by r! or just r?

for example,
3-2-2 <--- you can only multiply by 3
3-1-3 <--- 3
4-1-2 <--- 3!
5-1-1 <--- 3

I understand why they're different when i write them out. But if i were faced with a more complicated arrangement i might not be able to discern which to multiply...

[Goldust]

For arrangements with some items identical, there is a formula.

Suppose you have p items of which a items are identical, b other items are identical and so on till n identical items.

Then, the total number of arrangements is given by p!/(a!b!...n!)

In the first case (3-2-2), you have one 3 and two 2's. Hence, the total arrangements are 3!/2! = 3. Ditto for the second and fourth case where you have two 3's and two 1's respectively.

In the third case, you have 3 distinct elements and hence, the total arrangements are simply 3!

[mystic87]

wow thanks soo much! i dont believe i've ever learned that formula in highschool..

[mystic87]

from Nova:

Example 2:
Column A
The number of distinct
prime factors of x
Column B
The number of distinct
prime factors of 4x

We expect the number of prime factors of 4

x to be larger than the number of prime factors of x. But that is

the eye-catcher. Now, the number of prime factors of 4x cannot be less than the number of prime factors of

x since 4x contains all the factors of x. So the answer must be that either they are equal or there is not enough information. In fact, there is not enough information, as can be verified by plugging in the numbers

x = 2 and then x = 3.

How is it possible that A and B are equal?

[Goldust]

A and B could be equal if it is given that x is even.

[mystic87]

OHHHHH the number of distinct factor, not # of factors. i didn't realize that until i typed out all the factors. Just to make sure, if they actually reworded the question to "number of prime factors" B would always be greater right?