Question

There are $8$ balls of different colours. In how many ways can we select $5$ balls, so as to
(i) Include a white ball.
(ii) Exclude a red and black ball.

Answer 1

Here the given question is based on the concept of combination. We have to select 5 balls out of 8 balls given two different conditions. So for this, we will use the formula of combination i.e., calculating we have to select $'r'$ objects out of $'n'$ different objects gives $^n{C_r}$ . On substituting and simplifying the formula we get the required solution.

Complete step by step answer:
Combination is defined as “the arrangement of ways to represent a group or number of objects by selecting them in a set and forming the subsets”. Generally, combination denoted by $^n{C_r}$, \left( {\begin{array}{*{20}{c}} n \\\ r \end{array}} \right), or ‘$n$ choose $r$’.
The formula used to calculate the combination is: $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$----(1)
Where, ‘$n$’ is the number of items you have to choose from (total number of objects) and ‘$r$’ is the number of items you're going to select.
Now consider the question, given, the 8 balls of different colours. We have to find the number of ways to select 5 balls out of 8 balls by applying the following conditions:

(i) Include a white ball. Total number of balls is 8. If one ball is white which is already included, then we need to select 4 balls out of 7 balls. Therefore, the number of ways is $^7{C_4}$. Now, by using a formula,
${\,^7}{C_4} = \dfrac{{7!}}{{\left( {7 - 4} \right)!\, \cdot 4!}}$
$\Rightarrow \,\,\,\dfrac{{7!}}{{3!\, \cdot 4!}}$
$\Rightarrow \,\,\,\dfrac{{7 \times 6 \times 5 \times 4!}}{{\left( {3 \times 2 \times 1} \right) \cdot 4!}}$
On cancelling the like terms, then we have
$\dfrac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}}$
$\Rightarrow \,\,\,\dfrac{{210}}{6}$
On simplification, we get
$\therefore \,\,\,35$ ways

Thus, the number of ways in which 5 balls out of 8 such including a white ball is 35.

(ii) Exclude a red and black ball. Total number of balls is 8. If two balls red and black are not to be selected, then we need to select 5 balls out of 6 balls. Therefore, the number of ways is $^6{C_5}$. Now, by using a formula
${\,^6}{C_5} = \dfrac{{6!}}{{\left( {6 - 5} \right)!\, \cdot 5!}}$
$\Rightarrow \,\,\,\dfrac{{6!}}{{1!\, \cdot 5!}}$
$\Rightarrow \,\,\,\dfrac{{6 \times 5!}}{{5!}}$
On simplification, we get
$\therefore \,\,\,6$ Ways

Thus, the number of ways in which 5 balls out of 8 excluding a red and black ball is 6.

Note: In combinations each of the different selections made by taking some or all of a number of objects irrespective of their arrangement. Remember, factorial is the continued product of first n natural numbers is called the “n factorial “ and it is represented by $n! = \left( {n - 1} \right) \cdot \left( {n - 2} \right) \cdot \left( {n - 3} \right).....3 \cdot 2 \cdot 1$. We also know that if we have $0!$, then the value of this is $0! = 1$.