Combinations Problem

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.