Question

Find the number of ways of selecting $9$ balls from $6$red balls, $5$ white balls and $5$ blue balls if each selection consists of $3$ balls of each color.

Answer 1

Hint: Here we are asked to find the no. of ways nine balls can be selected so that each selection consists of three balls of each color. We will solve this problem by using a combination formula. First, we will find the combination of selecting three balls from each color then we will multiply them to get the result.
Formula: Formula that we need to know:
Combination: $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$n - $Total no. of items in the set
$r - $Required no. of item from the set (no. of items that we want to choose)

Complete step by step answer:
It is given that there are six red balls, five white balls, and five blue balls in total. We aim to find the no. of ways of selecting nine balls from this total so that it contains three balls of each color.
Since the requirement is to get three balls of each color, we will first find the combination of selecting three balls from each set. This can be done by using a combination formula that is
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
$n - $Total no. of items in the set and $r - $required no. of items from the set (no. of items that we want to choose).
Selecting three balls from six red balls will be $^6{C_3}$
Selecting three balls from five white balls will be $^5{C_3}$
Selecting three balls from five blue balls will be $^5{C_3}$
Thus, selecting nine balls from six red balls, five white balls, and five blue balls so that we get three balls of each color will be $^6{C_3}{ \times ^5}{C_3}{ \times ^5}{C_3}$
Expanding the above using the combination formula we get
$^6{C_3}{ \times ^5}{C_3}{ \times ^5}{C_3} = \dfrac{{6!}}{{3!3!}} \times \dfrac{{5!}}{{3!2!}} \times \dfrac{{5!}}{{3!2!}}$
On simplifying this we get
$ \Rightarrow \dfrac{{6 \times 5 \times 4 \times 3!}}{{3! \times 3 \times 2 \times 1}} \times \dfrac{{5 \times 4 \times 3!}}{{3! \times 2 \times 1}} \times \dfrac{{5 \times 4 \times 3!}}{{3! \times 2 \times 1}}$
On further simplification we get
$ \Rightarrow 5 \times 4 \times 5 \times 2 \times 5 \times 2$
$ \Rightarrow 2000$
Thus, there are two thousand ways to get nine balls of each color from six red, five white, and five blue balls.

Note:
In this problem, we first found the combination of getting three balls from each set of colors but we aimed to find the combination of getting nine balls with three balls of each color so we have used the multiplication operation to combine them. Sometimes students get confused between addition and multiplication like which operation we should use to solve.