A class contains \[4\] boys and $g$ girls. Every Sunday five students, including at least three boys go for a picnic to zoo park, a different group being sent every week. During the picnic, the class teacher gives each girl in the group a doll. If the total number of dolls distributed was $85$, then the value of g is.
\[\begin{array}{*{20}{l}}
  {\left( A \right){\text{ }}15} \\
  {\left( B \right){\text{ }}12} \\
  {\left( C \right){\text{ }}8} \\
  {\left( D \right){\text{ }}5}
\end{array}\]

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Hint:In the given question we have to find out the value of ‘g’ , First we have to find out the cases formed when $3$ boys or \[4\] boys are selected for the picnic. After that, find out the number of dolls distributed per case. Since the total number of dolls are given, Equate the number of dolls found out with the total no. of dolls given.

Complete step-by-step answer:
In a class, there are \[4\] boys and g girls Each Sunday only $5$ students are allotted to go for pieties in which at least $3$ are boys.
From given information’s two cases are formed –
According to 1st – case –
Number of boys is $3$.
Then, the number of girls is $2$.
According to 2nd case given in the question –
Number of boys are \[4\].
Then, the number of girls is $1$.
As, $3$ boys are selected in 4 boys and
$2$ girls are selected in g girls and 2 dolls are for 2 girls as per question, a doll is given to each girl.
No. of dolls in first case –
$ = {\;^4}{C_3}{ \times ^8}{C_2} \times 2 - - - - - - - (i)$

As, $4$ boys are selected in 4 boys and
$1$ girl is selected in g girls and only one girl is present, so the number of dolls becomes 1.
No. of dolls in 2nd case –
$ = {\;^4}{C_4}{ \times ^g}{C_1} \times 1 - - - - - - (ii)$

So, the total no. of dolls distributed is \[85\].
$\therefore {\;^4}{C_3}{ \times ^g}{C_2}{ \times ^4}{C_4}{ \times ^g}{C_1} \times 1 = 85.$
$ \Rightarrow \dfrac{{4!}}{{3! \times 1!}} \times \dfrac{{g!}}{{2! \times \left( {g - 2} \right)!}} \times \dfrac{{4!}}{{4! \times 0!}} \times \dfrac{{g!}}{{\left( {g - 1} \right)! \times 11}} \times 1 = 85$
$ \Rightarrow 4 \times \dfrac{{g!}}{{2! \times \left( {g - 2} \right)!}} \times 2 + \dfrac{{g!}}{{\left( {g - 1} \right)!}} = 85$
\[ \Rightarrow 8 \times \dfrac{{g!}}{{2! \times \left( {g - 2} \right)!}} + \dfrac{{g!}}{{\left( {g - 1} \right)\left( {g - 2} \right)!}} = 85\left[ {\because \left( {n - n} \right)! = \left( {n - n} \right)\left( {n - n - 1} \right)!} \right]\]
\[ \Rightarrow \dfrac{{g!}}{{\left( {g - 2} \right)!}}\left[ {4 + \dfrac{1}{{g - 1}}} \right] = 85\] [ Simplifying the equation]
$ \Rightarrow \dfrac{{g\left( {g - 1} \right)\left( {g - 2} \right)!}}{{\left( {g - 2} \right)!}} \times \dfrac{{\left( {4g - 3} \right)}}{{\left( {g - 1} \right)}} = 85$ [ by cancellation of terms present both as numerator & denominator]
$ \Rightarrow g\left( {4g - 3} \right) = 85$ [ simplifying the equation]
$ \Rightarrow 4{g^2} - 3g - 85 = 0$ [ By middle term factorisation]
$ \Rightarrow 4{g^2} - 20g + 17g - 85 = 0$ [ solving the equation for ‘g’]
$\therefore \left( {4g + 17} \right)\left( {g - 5} \right) = 0$
Either,
     $g = - \dfrac{{17}}{4}\left( {not\;possible} \right)$ or \[g{\text{ }} = {\text{ }}5\]\[g{\text{ }} = {\text{ }}5\]
$\therefore\,{\text{value of g is 5}}$

So, the correct answer is “Option D”.

Note: In the solution, you can also find the answer by putting the value given in the option, in the expression –
$\dfrac{{g!}}{{\left( {g - 2} \right)!}}\left[ {4 + \dfrac{1}{{g - 1}}} \right] = 85$. If you don’t want, then solve the expression by using the middle term as in the solution.The term permutation is used where we have to arrange some object in different ways and combination is used when we have to choose something from a given object.