Question

Among a set of \[5\] black balls and \[3\] red balls. How many selections of \[5\] balls can be made such that at least \[3\] of them are black balls?

Answer 1

Hint: We have to use the combinations formula to solve this question. Combinations formula is \[{}^n{C_r} = \dfrac{{n!}}{{[r!(n - r)!]}}\]

Complete step-by-step solution:
Based on the question above we were able to identify that there is a set of \[5\] black balls and \[3\] red balls.
And we need to find the selection of \[5\] balls which is to be selected randomly, among that selection there should be at least \[3\] balls that should be black in color.
That is,
Selecting at least \[3\] black balls from a set of \[5\] black balls in a total selection of \[5\] balls can be mentioned as follows,
Step 1: The set can be as consists of \[3\] black balls and \[2\] red balls.
While expressing it in the form of combination it’s as \[{}^5{C_3} \times {}^3{C_2}\]
Step 2: Another set may consist of \[4\] black balls and \[1\] red balls.
Similarly, while expressing the above selection set in the combination form as \[{}^5{C_4} \times {}^3{C_1}\]
Step 3: The next set may consist of \[5\] black balls and \[0\] red balls.
With this continuous format, while representing it into combination form as \[{}^5{C_5} \times {}^3{C_0}\]
Therefore, while the solution occurs in the format as combination of the form
Thus, our solution expression looks like,
By adding all the three set of combinations,
\[{}^5{C_3} \times {}^3{C_2}\]\[ + \]\[{}^5{C_4} \times {}^3{C_1}\]\[ + \]\[{}^5{C_5} \times {}^3{C_0}\]
Calculating the combination value for each of the term by this formula,
\[{}^n{C_r} = \dfrac{{n!}}{{[r!(n - r)!]}}\] ----------(1)
Step 1:
Calculating the combination value \[{}^5{C_3}\] and \[{}^3{C_2}\] by equation (1), then
For \[{}^5{C_3}\], since $n = 5,r = 3$
\[ \Rightarrow {}^5{C_3} = \dfrac{{5!}}{{[3!(5 - 3)!]}} = \dfrac{{5!}}{{[3! \times 2!]}}\]
On simplifying the numerator and denominator as follows, we get
\[\dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{[(3 \times 2 \times 1) \times (2 \times 1)]}} = \dfrac{{120}}{{[6 \times 2]}}\]
Therefore, \[{}^5{C_3} = \dfrac{{120}}{{12}} = 10\]
For \[{}^3{C_2}\], since $n = 3,r = 2$
\[ \Rightarrow {}^3{C_2} = \dfrac{{3!}}{{[2!(3 - 2)!]}} = \dfrac{{3!}}{{[2! \times 1!]}}\]
On simplifying the numerator and denominator as follows, we get
\[{}^3{C_2} = \dfrac{{3 \times 2 \times 1}}{{[(2 \times 1)(1)]}}\]
Therefore, \[{}^3{C_2} = \dfrac{6}{2} = 3\]
Step 2:
Calculating the combination value \[{}^5{C_4}\] and \[{}^3{C_1}\] by equation (1), then
For \[{}^5{C_4}\], since $n = 5,r = 4$
\[ \Rightarrow {}^5{C_4} = \dfrac{{5!}}{{[4!(5 - 4)!]}} = \dfrac{{5!}}{{[4! \times 1!]}}\]
On simplifying the numerator and denominator as follows, we get
\[ = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{[(4 \times 3 \times 2 \times 1)(1)]}} = \dfrac{{120}}{{[24 \times 1]}}\]
Therefore, \[{}^5{C_4} = \dfrac{{120}}{{24}} = 5\]
For \[{}^3{C_1}\], since $n = 3,r = 1$
\[ \Rightarrow {}^3{C_1} = \dfrac{{3!}}{{[1!(3 - 1)!]}} = \dfrac{{3!}}{{[1! \times 2!]}}\]
On simplifying the numerator and denominator as follows, we get
\[ = \dfrac{{3 \times 2 \times 1}}{{[(1) \times (2 \times 1)]}}\]
Therefore, \[{}^3{C_1} = \dfrac{6}{2} = 3\]
Step 3:
Calculating the combination value \[{}^5{C_5}\] and \[{}^3{C_0}\] by equation (1), then
For \[{}^5{C_5}\], since $n = 5,r = 5$
\[ \Rightarrow {}^5{C_5} = \dfrac{{5!}}{{[5!(5 - 5)!]}} = \dfrac{{5!}}{{[5! \times 1]}}\]
On simplifying the numerator and denominator as follows, we get
\[ = \dfrac{{5!}}{{5!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1}}\]
Therefore, \[{C_5} = \dfrac{{120}}{{120}} = 1\]
For \[{}^3{C_0}\], since $n = 3,r = 0$
As \[0! = 1\]
\[ \Rightarrow {}^3{C_0} = \dfrac{{3!}}{{[0!(3 - 0)!]}} = \dfrac{{3!}}{{[1 \times 3!]}}\]
On simplifying the numerator and denominator as follows, we get
\[ = \dfrac{{3!}}{{3!}} = \dfrac{{3 \times 2 \times 1}}{{3 \times 2 \times 1}}\]
Therefore, \[{}^3{C_0} = \dfrac{6}{6} = 1\]
After finding the values of all the combination forms need to be substitute the values of them in the expression follows,
\[ \Rightarrow \]\[{}^5{C_3} \times {}^3{C_2}\]\[ + \]\[{}^5{C_4} \times {}^3{C_1}\]\[ + \]\[{}^5{C_5} \times {}^3{C_0}\] \[ = (10 \times 3) + (5 \times 3) + (1 \times 1)\]
\[ = (30) + (15) + (1)\]
Therefore, the sum of the three set of combination: \[{}^5{C_3} \times {}^3{C_2}\]\[ + \]\[{}^5{C_4} \times {}^3{C_1}\]\[ + \]\[{}^5{C_5} \times {}^3{C_0}\]\[ = 46\]
So, for the given set of combinations, it can be expressed as in that of \[46\] ways.
\[46\] ways of selection of \[5\] balls can be made such that at least \[3\] of them are black balls.

Note: The important combination formulas are,
1. \[{}^n{C_n} = 1\]
2. \[{}^n{C_0} = 1\]
3. \[{}^n{C_1} = n\]
For the given question as per the need to find the selection set of \[5\] balls which is needed to be in the combination of at least \[3\] black balls from the set of \[5\] black balls and \[3\] red balls.
So based on the expression formation as, \[{}^5{C_3} \times {}^3{C_2}\]\[ + \]\[{}^5{C_4} \times {}^3{C_1}\]\[ + \]\[{}^5{C_5} \times {}^3{C_0}\], we obtain required solution.