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Halle, Julia and Drew have 5 donuts to share. If one of them [#permalink]
13 Oct 2004, 21:04

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Question Stats:

100% (01:23) correct
0% (00:00) wrong based on 1 sessions

Halle, Julia and Drew have 5 donuts to share. If one of them can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed. Answer:21

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:
C(n+r-1,r)
n number of elements
r-combinations.

So
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.
So,
*****||
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

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