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

Hint: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 \[0! = 1\].