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# 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?

Last updated date: 22nd Jul 2024
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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.