Question

A teacher takes 3 children from her class to the zoo at a time as often as she can, but she does not take the same three children to the zoo more than once. She finds that she goes to the zoo 84 times more than a particular child goes to the zoo. The number of children in her class is
A.12
B.10
C.60
D.None of these

Answer 1

Hint: First we’ll find the total number of days the teacher has gone to the zoo by finding the total number of ways of selecting 3 students out of the total students, we’ll find the number of days a particular student has visited the zoo.
Now we have given that the teacher goes 84 times more than a particular student, using this we’ll create and solve the equation to get the total strength of the students.

Complete step-by-step answer:
We know that number of ways selecting or pairing ‘r’ elements out of the total number of ‘n’ elements is given by ${}^n{C_r}$ which is mathematically equal to $\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
i.e. ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Let the total number of students are ’n’
Since the teacher goes with every group of three students, the number of different groups are ${}^n{C_3}$
The number of ways that a particular student goes to the zoo i.e. selecting any 2 students out of remaining \[\left( {n - 1} \right)\] students i.e. ${}^{n - 1}{C_2}$
It is given that the teacher goes 84 times more than a particular student i.e. \[{}^n{C_3} - {}^{n - 1}{C_2} = 84\]
Using ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ ,
\[ \Rightarrow \dfrac{{n!}}{{3!\left( {n - 3} \right)!}} - \dfrac{{(n - 1)!}}{{2!\left( {n - 1 - 2} \right)!}} = 84\]
Using $n! = n(n - 1)!$ , we get
\[ \Rightarrow \dfrac{{n(n - 1)(n - 2)(n - 3)!}}{{3!\left( {n - 3} \right)!}} - \dfrac{{(n - 1)(n - 2)(n - 3)!}}{{2!\left( {n - 3} \right)!}} = 84\]
On simplifying we get,
\[ \Rightarrow \dfrac{{n(n - 1)(n - 2)}}{{3!}} - \dfrac{{(n - 1)(n - 2)}}{{2!}} = 84\]
Taking \[\left( {n - 1} \right)\left( {n - 2} \right)\] common we get,
\[ \Rightarrow \left( {n - 1} \right)\left( {n - 2} \right)\left( {\dfrac{n}{{3!}} - \dfrac{1}{{2!}}} \right) = 84\]
Simplifying the brackets by taking LCM we get,
\[ \Rightarrow \left( {n - 1} \right)\left( {n - 2} \right)\left( {\dfrac{{n - 3}}{{3!}}} \right) = 84\]
Multiplying both sides by \[3!\] i.e. 6
\[ \Rightarrow \left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 84 \times 6\]
On splitting the RHS into factors we get,
\[ \Rightarrow \left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) = 9 \times 8 \times 7\]
On comparing we can say that
\[ \Rightarrow \left( {n - 1} \right) = 9\]
\[\therefore n = 10\]
Option(B) is correct .

Note: While finding the number of days that a particular student went to the zoo most of the student take the possible ways as ${}^n{C_2}$ , they do not exclude the student from the total students which given them incorrect answer, so remember that we have excluded that student from the total number of students as he has already been selected.