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MBA.Aspirant
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I have a feeling the question means to ask the following: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?
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.
