GMATinsight wrote:
In how many ways can 6 chocolates be distributed among 3 children?
A child may get any number of chocolates from 0 to 6 and all the chocolates are identical.A) 21
B) 28
C) 56
D) 112
E) 224
SOURCE: http://www.GMATinsight.comLet, three children get a, b and c chocolates respectively
i.e. a+b+c = 6
Now, the whole number solutions of an equation = (n+r-1)C(r-1) = (6+3-1)C(3-1) = 8C2 = 28
Answer: Option B
ALTERNATIVELet, three children get a, b and c chocolates respectively
i.e. a+b+c = 6
@a=0, (b,c) can be (0,6) (1,5) (2,4) (3,3) (4,2) (5,1) (6,0) i.e. 7 ways
@a=1, (b,c) can be (0,5) (1,4) (2,3) (3,2) (4,1) (5,0) i.e. 6 ways
@a=2, (b,c) can be (0,4) (1,3) (2,2) (3,1) (4,0) i.e. 5 ways
and so on...
Total Ways of distribution = 7+6+5+4+3+2+1 = 28
Answer: Option B
To attend the !!!FREE!!! Live class on Permutation and combination on 24th July 2019 @9 PM IST, click and register on the LINK.
;
Hello, can you please explain why the whole number solutions of the equation = (n+r-1)C(r-1)?
close
52% (01:27) correct 
