How many balls can be selected?

Question

A sack contain 20 identical red balls, 20 identical blue balls, and 20 identical green balls. In how many distinct ways can 10 balls be selected from the sack?

Answer 1

Because we only have $3$ colours, we can imagine making an organized list of possibilities, and count them relatively simply. We organize the list by using the number of red balls, which can be $10$, $9$, $8$, and so on down to $0$.

i) $10$ red: There is only $1$ way to have this.

ii) $9$ red: We could have $1$ blue, or $0$ blue, $2$ possibilities. For note that once we have decided on the number of red and the number of blue, the number of green is determined.

iii) $8$ red: We could have $2$ blue, or $1$, or $0$, a total of $3$ possibilities.

iv) $7$ red: We could have $3$ blue, or $2$, or $1$, or $0$, a total of $4$ possibilities.

v) $6$ red: The same reasoning as above shows there are $5$ possibilities.

vi) $5$ red: There are $6$ possibilities.

vii) $4$ red: There are $7$ possibilities.

The remaining cases, from $3$ red to $0$ red, give $8$ to $11$ possibilities.

Thus the total is $1+2+3+\cdots +11$. We can add slowly, or use the formula for the sum of an arithmetic progression. That gives a total of $\frac{(11)(12)}{2}$.

Another way: Let $x_1$ be the number of red, $x_2$ the number of blue, and $x_3$ be the number of green. We want to find the number of solutions of $x_1+x_2+x_3=10$ in non-negative integers. By the Stars and Bars Method, this number is $\binom{10+3-1}{10}$.

Answer 2

There are $66$ distinct ways to select 10 balls from 3 sets of 20 identical balls.

$$\sum_{r=0}^{10}\sum_{b=0}^{10-r} 1 = \sum_{r=0}^{10} (11-r) = 121-\frac{10(10+1)}{2} = 66$$