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Combinations Problem

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Combinations Problem

by serenayc » Sun May 06, 2012 8:33 pm
L, M and D have 5 donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?

ANS: 21
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by aneesh.kg » Sun May 06, 2012 8:41 pm
Interesting.
Note that the donuts are identical. So we are only concerned with who gets how many donuts.

Listing down all the possible distribution on number of donuts among three people and then arranging it among L, M and D.
0-0-5: 3!/2! = 3 (because there are 2 0s, we divide 3! by 2!)
0-1-4: 3! = 6
0-2-3: 3! = 6
1-1-3: 3!/2! = 3
1-2-2: 3!/2! = 3

Adding all of them, the total number of ways = 3 + 6 + 6 + 3 + 3= 21

Note that the three ways of arranging 0-0-5 are 0-0-5, 0-5-0 and 5-0-0,
and
the six ways of arranging 0-1-4 are 0-1-4,0-4-1,4-0-1,4-1-0,1-4-0,1-0-4,
and so on.
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