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Question: Find the number of ways of selecting \(9\) balls from \(6\)red balls, \(5\) white balls and \(5\) bl...

Find the number of ways of selecting 99 balls from 66red balls, 55 white balls and 55 blue balls if each selection consists of 33 balls of each color.

Explanation

Solution

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: nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
n−n - Total no. of items in the set
r−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
nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
n−n - Total no. of items in the set and r−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 6C3^6{C_3}
Selecting three balls from five white balls will be 5C3^5{C_3}
Selecting three balls from five blue balls will be 5C3^5{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 6C3×5C3×5C3^6{C_3}{ \times ^5}{C_3}{ \times ^5}{C_3}
Expanding the above using the combination formula we get
6C3×5C3×5C3=6!3!3!×5!3!2!×5!3!2!^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
⇒6×5×4×3!3!×3×2×1×5×4×3!3!×2×1×5×4×3!3!×2×1\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
⇒5×4×5×2×5×2\Rightarrow 5 \times 4 \times 5 \times 2 \times 5 \times 2
⇒2000\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.