Question

A teacher takes three children from her class to the zoo at a time, but she doesn't take the same three children to the zoo more than once. She finds that she went to the zoo 84 times. More than a particular child has gone to the zoo. The number of children in her class is
A. 12
B. 10
C. 60
D. None of them.

Answer 1

Hint: First of all we will suppose the total number of children present in the class to be 'n'. We know that the number of ways of selecting 'r' things among n different things is equal to ${}^n{C_r}$ . So, we will use this to find the required value of n.

Complete step-by-step solution:
We have been given that a teacher takes three children from her class to the zoo at a time, but she doesn't take the same three children to the zoo more than once. Also, she went to the zoo 84 times. More than a particular child.
Let the number of children in the class be 'n'.
We know that the number of ways a teacher can go to the zoo is the number of ways a group of 3 children can be made from n children.
Selecting 3 children from n children $ = {}^n{C_3}$
Number of ways a particular child is chosen = Number of ways 2 children can be selected from remaining $\left( {n - 1} \right)$ children $ = {}^{n - 1}{C_2}$
According to the question,
${}^n{C_3} - {}^{n - 1}{C_2} = 84$
We know that ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\, \times r!}}$
\[\begin{array}{l} \Rightarrow \dfrac{{n!}}{{\left( {n - 3} \right)!\, \times 3!}} - \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 3} \right)!\, \times 2!}} = 84\\ \Rightarrow \dfrac{{\left( {n - 3} \right)!\left( {n - 2} \right)\left( {n - 1} \right)n}}{{\left( {n - 3} \right)!\, \times 6}} - \dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)!}}{{\left( {n - 3} \right)!\, \times 2}} = 84\\ \Rightarrow \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6} - \dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)}}{2} = 84\end{array}\]
Taking $LCM$ of the terms, we get:
\[\dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right) - 3\left( {n - 1} \right)\left( {n - 2} \right)}}{6} = 84\]
Taking out the common terms, we get:
\[\begin{array}{l}\dfrac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{6} = 84\\\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 84 \times 6 = 12 \times 7 \times 6\\ = 4 \times 3 \times 7 \times 3 \times 2\\ = 9 \times 8 \times 7\end{array}\]
On comparing both sides, we get
$ \Rightarrow n = 10$
Therefore, the correct option is B.

Note: Sometimes by mistake we misunderstand the question that 84 is the way a teacher can go to the zoo and equate ${}^n{C_3}$ is equal to 84. So, we get the incorrect answer.
Also, we can use the option to find the value of 'n' by substituting the option in the equation formed as it will save your time.