I think the answer should be 21bomond wrote:Larry, Michael, and Doug have five 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?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
i would go with 21 as well..bomond wrote:Larry, Michael, and Doug have five 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?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
sudhir3127 wrote:i would go with 21 as well..bomond wrote:Larry, Michael, and Doug have five 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?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
formula is
"distributing n identical things among r persons such that any person might get any number ( incl 0) is n+r -1 C r-1"
thus its
5+3-1 C 3-1
7C2 = 21
Hope that helps..
C stands combinationroman.savchenko wrote: what is C between 7 and 2? thanks
sudhir3127, could you please specify? I see such formula at first timesudhir3127 wrote: formula is
"distributing n identical things among r persons such that any person might get any number ( incl 0) is n+r -1 C r-1"
thus its
5+3-1 C 3-1
7C2 = 21
the formula for4meonly wrote:sudhir3127, could you please specify? I see such formula at first timesudhir3127 wrote: formula is
"distributing n identical things among r persons such that any person might get any number ( incl 0) is n+r -1 C r-1"
thus its
5+3-1 C 3-1
7C2 = 21
I solved this problem in such way:
005 = 3!/2! = 3
104 = 3!/1 = 6
113 = 3!/2! = 3
230 = 3!/1 = 6
221 = 3!/2! = 3
adding 3+6+3+6+3=21
But your's approach is faster.
I will highly appreciate your comments about formula
n+r -1 C r-1
what about when zero is not included?
If this problem will look like
Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 1 to 3, in how many different ways can the donuts be distributed?
Will this formala work?
I think in this question answer will be 3 ways
I doubt you'll ever need this formula on the GMAT, but if you're interested in how to do the problem (for any number of donuts) without a formula, or if you want to know why the formula works:sudhir3127 wrote:
i would go with 21 as well..
formula is
"distributing n identical things among r persons such that any person might get any number ( incl 0) is n+r -1 C r-1"
What is the non-formula approach you are talking about Ian ? And please let us know the BEST approach for such ( Perm/Comb) Problems.Ian Stewart wrote: I doubt you'll ever need this formula on the GMAT
mmm.. donuts...4meonly wrote:6 ways?:
113
131
311
I have only 3 ways
3!/2! = 3
I can find only 6 places...Ian Stewart wrote: | O O O O | O
If you count, we have seven slots in which we could place the two partition markers (the '|' symbols), and the order in which we place them does not matter:
