Same but not so same

Let $x_{i}$ be the number of balls in box $i$ . Then we have,

$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=6$ .

Thus, using stars and bars , there are $^{11}C_{5}=\boxed{462}$ ways.

The balls are indistinguishable, but the boxes are not. Thus (0,0,0,0,0,6) is different from (6,0,0,0,0,0). Basically, we are trying to find the number of 6-tuples. We can do this by finding the number of combinations for 11111100000. There are six 1's representing the six balls and five 0's representing seperaters. For example a (1,3,0,0,2,0) would be 10111000110, a (1,1,1,1,1,1) would be a 10101010101 and so on. You can better visualize this be replacing each comma in a tuple with a 0. The number of ordered tuples would then by 11C6 which is $\boxed{462}$

This is a simple case of stars and bars or sticks and stones or whatever you call it. We have ||||| ** where | represents a divider between boxes and * represents a ball. This is $\binom{5+6}{5} = \boxed{462}$ .

A+B+C+D+E+F=6 where A,B,C ... are the boxes.... Now it becomes simple stars and bars problem; i.e Nu of ways = (6+5) \times C {5}=11 \times C {5}=\boxed{462}

11c5

Existem cinco "espaços", entre as caixas, e existem as 6 bolas. Portanto:

\frac{(6+5)!}{5! \times 6!} = 462

6 balls are to be divided amongst 6 boxes,derfore according to formula 6+6-1c6-1=>11c5=462

if there are N things and it has to be distributed between R no. of peoples then total no of ways

=combination(N+R-1 , R-1)=C( 11 , 5 ) =462