Question

In a class there are 10 boys and 8 girls. The teacher wants to select either a boy or a girl to represent the class in a function. Find the number of ways the teacher can make this selection.

Answer 1

Hint: Here, first we have to select one boy from 10 boys. The number of possible ways is ${}^{10}{{C}_{1}}$. Next, we have to select one girl from 8 girls. The number of possible ways is ${}^{8}{{C}_{1}}$. At last we have to calculate the number of ways of selecting either a boy or a girl, which is the product of number of possible ways of selecting a boy and number of possible ways of selecting a girl.

Complete step-by-step answer:
We are given that in a class there are 10 boys and 8 girls. The teacher wants to select either a boy or a girl to represent the class in a function.
Now, we have to find the number of ways the teacher can make this selection.
We have a total of 10 boys, from 10 we have to select a boy. Therefore, we can write:
The number of ways to select a boy = ${}^{10}{{C}_{1}}$
We have a formula that:
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Hence, by applying the above formula we will get:
 $\begin{align}
  & {}^{10}{{C}_{1}}=\dfrac{10!}{1!\left( 10-1 \right)!} \\
 & {}^{10}{{C}_{1}}=\dfrac{10!}{1!9!} \\
\end{align}$
We have:
$\begin{align}
  & 1!=1 \\
 & 9!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9 \\
 & 10!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10 \\
\end{align}$
By substituting the above values we obtain:
${}^{10}{{C}_{1}}=\dfrac{1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9\times 10}{1\times 1\times 2\times 3\times 4\times 5\times 6\times 7\times 8\times 9}$
Now, by cancellation we obtain:
${}^{10}{{C}_{1}}=10$
We also have a total of 8 girls, from 8 we have to select a girl. Hence we can write:
The number of ways to select a girl = ${}^{8}{{C}_{1}}$
$\begin{align}
  & {}^{8}{{C}_{1}}=\dfrac{8!}{1!\left( 8-1 \right)!} \\
 & {}^{8}{{C}_{1}}=\dfrac{8!}{1!7!} \\
\end{align}$
We have:
$\begin{align}
  & 1!=1 \\
 & 7!=1\times 2\times 3\times 4\times 5\times 6\times 7 \\
 & 8!=1\times 2\times 3\times 4\times 5\times 6\times 7\times 8 \\
\end{align}$
Next, by substituting all these values we obtain:
${}^{8}{{C}_{1}}=\dfrac{1\times 2\times 3\times 4\times 5\times 6\times 7\times 8}{1\times 1\times 2\times 3\times 4\times 5\times 6\times 7}$
Now, by cancellation we get:
${}^{8}{{C}_{1}}=8$
The number of ways to select either a boy or a girl = The number of ways to select a boy $\times $ The number of ways to select a girl
Therefore, we will get:
The number of ways to select either a boy or a girl = $10\times 8$
i.e. the number of ways to select either a boy or a girl = 80
Hence, we can say that there are 80 possible ways to select either a boy or a girl.

Note: Here, we have to select either a boy or a girl from the group of 10 boys and 8 girls, so, we should find the product of the number of ways of selecting a boy and selecting a girl. That means here, “either” “or” is used, in such cases we always have to find the product.