MBA.Aspirant wrote:This morning Pat went to the market with a red bag and a green one. If he bought 10 carrots and 6 radishes, in how many ways could he distribute these vegetable so that no bag would be empty?

I have a feeling the question means to ask the following:

This morning Pat went to the market with a red bag and a green one. If he bought 10 IDENTICAL carrots and 6 IDENTICAL radishes, in how many ways could he distribute these vegetables so that no bag would be empty?

We can use the separator method.

Let CCCCCCCCCC = the 10 identical carrots and RRRRRR = the 6 identical radishes.
Let | = separator.

We need one separator to represent how the the vegetables can be divided between the two bags.
CCCCC|CCCCC = 5 carrots in each bag.
CC|CCCCCCCC = 2 carrots in one bag, 8 in the other.
CCCCCCC|CCC = 7 carrots in one bag, 3 in the other.

To count all the possible distributions, we need to count the number of ways to arrange CCCCCCCCCC| and RRRRRR|.
Any arrangement of these elements represents a possible distribution of the two vegetables.
We must account for the identical elements, which reduce the number of unique arrangements:

Number of ways to arrange N elements, of which S elements are identical = N!/S!.

Given CCCCCCCCCC|, we are arranging 11 elements, of which 10 are identical:
Number of ways to arrange CCCCCCCCCC| = 11!/10! = 11.

Given RRRRRR|, we are arranging 7 elements, of which 6 are identical:
Number of ways to arrange RRRRRR| = 7!/6! = 7.

To combine the number of carrot distributions with the number of radish distributions, we multiply the results above:
11*7 = 77.

Two of the these distributions are not allowed: empty red bag and full green bag, full red bag and empty green bag.
Thus, we must subtract these two bad distributions:
77-2 = 75.

MBA.Aspirant wrote:This morning Pat went to the market with a red bag and a green one. If he bought 10 carrots and 6 radishes, in how many ways could he distribute these vegetable so that no bag would be empty?

may i ask where you find this question? because its way to long to know the answer and time consuming question never exists in GMAT according to what i know... if you find this question in such books as manhatan, kaplan, master or OG, i think i should be worried...

diebeatsthegmat wrote:

MBA.Aspirant wrote:This morning Pat went to the market with a red bag and a green one. If he bought 10 carrots and 6 radishes, in how many ways could he distribute these vegetable so that no bag would be empty?

may i ask where you find this question? because its way to long to know the answer and time consuming question never exists in GMAT according to what i know... if you find this question in such books as manhatan, kaplan, master or OG, i think i should be worried...

A more GMAT-friendly way to solve this problem:

All we need to do is count the number of choices for the green bag.
This will account for all of the choices for the red bag, since any vegetable not in the green bag must automatically be in the red bag.

Green bag:
Number of carrots could be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10 = 11 choices.
Number of radishes could be 0, 1, 2, 3, 4, 5, or 6 = 7 choices.
To combine the number of choices we have for each vegetable, we multiply:
11*7 = 77 choices.

Since neither of the bags can be empty, 2 of these 77 choices are not allowed: all of the vegetables in the green bag and none of the vegetables in the green bag.
Subtracting the bad choices, we get:
77-2 = 75 possible distributions.