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Just have to think about it the right way. Basically we have a bag of donuts from which we will pull 5 times. Any donut removed from the bag can be a Halle, Julia, or Drew Donut.
The general formula is:
n number of elements
C(5+3-1,5) = C(7,5) = C(7,2) = 21
The way that this makes sense is if we consider aligning the 5 donuts (*) and placing separators between them to indicate to whom they belong.
would indicate that Halle has all the donuts,
would indicate that Julia has all the donuts, and
would indicate that Drew has all the donuts.
So the answer is the number of ways to pick the 5 *'s or the 2 |'s out of the set of 7.
Extending this logic further can also be used to solve problems like these:
1. In how many ways can 20 be expressed as sum of 5 non negative integers.
using similar logic : you have 20 *s are 4 seperators '|' and u have to chose 4 '|' and u can do that in 24C4 ways.
TO add to this twist, in how many ways can the number 20 be written as a sum of 5 positive integers? you have 20 *'s and u have to place 4 '|' in between them, i.e. no of ways in which 4 '|' can be placed in 19 places, where all '|' are identical. 19C4.
To summarise, no of ways in which n things can be divided among r persons so that each of them can receive 0 or more is n+r-1Cr-1. In tapsemi's Q. n= 5, r = 3, so reqd ways = 7C2 = 21.
and no of ways in which n things can be divided among r persons so that each of them receive at least 1 is n-1Cr-1