Question

A drawer contains a mixture of red marbles and blue marbles, at most 17 in all. It so happens that when two marbles are selected randomly without replacement, there is a probability of exactly $\dfrac{1}{2}$ that both are either red or blue. Let n be the largest possible number of red marbles in the drawer, consistent with the data. Then $n - 10$ equals

Answer 1

Hint : Let the no. of red marbles be x and the no. of blue marbles be y. Total number of marbles will be $x + y$ which is less than or equal to 17. Use combinations to select two marbles from x and y as the order of selection does not matter. Find the probability when two marbles are selected randomly without replacement from x or y and equate it to $\dfrac{1}{2}$ . Use this info to further solve the question.

Complete step-by-step answer :
We are given that a drawer contains at most 17 red and blue marbles. The probability when two marbles are red or two marbles are blue when selected randomly is $\dfrac{1}{2}$ .
We have to find the number of red marbles say n and the value of $n - 10$
Let the no. of red marbles be x and no. of blue marbles be y.
Which means that $x + y$ is the total no. of marbles and $x + y \leqslant 17$
No. of ways of selecting 2 red balls randomly is ${}^x{C_2}$ and no. of ways of selecting 2 blue balls randomly is ${}^y{C_2}$ ; selecting 2 balls randomly from total no. of balls is ${}^{x + y}{C_2}$
Probability when two marbles are red or two marbles are blue when selected randomly is
$\dfrac{{{}^x{C_2} + {}^y{C_2}}}{{{}^{x + y}{C_2}}}$ which is equal to $\dfrac{1}{2}$
$\Rightarrow \dfrac{{{}^x{C_2} + {}^y{C_2}}}{{{}^{x + y}{C_2}}} = \dfrac{1}{2} \\\ {}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} \\\ {}^x{C_2} = \dfrac{{x\left( {x - 1} \right)}}{2},{}^y{C_2} = \dfrac{{y\left( {y - 1} \right)}}{2},{}^{x + y}{C_2} = \dfrac{{\left( {x + y} \right)\left( {x + y - 1} \right)}}{2} \\\ \Rightarrow \dfrac{{\left( {\dfrac{{x\left( {x - 1} \right)}}{2}} \right) + \left( {\dfrac{{y\left( {y - 1} \right)}}{2}} \right)}}{{\left( {\dfrac{{\left( {x + y} \right)\left( {x + y - 1} \right)}}{2}} \right)}} = \dfrac{1}{2} \\\ \Rightarrow 2\left( {\dfrac{{x\left( {x - 1} \right) + y\left( {y - 1} \right)}}{2}} \right) = \dfrac{{\left( {x + y} \right)\left( {x + y - 1} \right)}}{2}{\text{(}}\therefore {\text{Cross multiplication)}} \\\ \Rightarrow {\text{2}}\left( {{x^2} - x + {y^2} - y} \right) = {x^2} + xy - x + xy + {y^2} - y \\\ \Rightarrow 2{x^2} - 2x + 2{y^2} - 2y = {x^2} + {y^2} + 2xy - x - y \\\ \Rightarrow 2{x^2} - {x^2} + 2{y^2} - {y^2} - 2xy = - x + 2x - y + 2y \\\ \Rightarrow {x^2} + {y^2} - 2xy = x + y \\\ \therefore {\left( {x - y} \right)^2} = x + y \;$
We already know that $x + y \leqslant 17$
Substitute the value of $x + y$ obtained in the above inequation.
$\left( {x + y} \right) \leqslant 17 \\\ \Rightarrow {\left( {x - y} \right)^2} \leqslant 17 \;$
As $x + y$ is a count, it must be a positive integer.
Which means the square root of ${\left( {x - y} \right)^2}$ must be a positive integer. So the greatest perfect square below 17 is our answer.
${\left( {x - y} \right)^2} \leqslant 17 \\\ \Rightarrow {\left( {x - y} \right)^2} = 16 \\\ \Rightarrow {\left( {x - y} \right)^2} = {4^2} \\\ \Rightarrow x - y = 4 \;$
On substituting the value of $x - y$ in ${\left( {x - y} \right)^2} = x + y$ , we get
$x + y = {4^2} \\\ \Rightarrow x + y = 16 \\\ \Rightarrow x = 10,y = 6 \;$
Total no. of red marbles x, which is n, is 10.
Then $n - 10 = 10 - 10 = 0$

Note : A Permutation is arranging the objects in order. Combination is the way of selecting an object from a group of objects. When the order of the objects does not matter then it should be considered as Combination and when the order matters then it should be considered as Permutation. Do not confuse using a combination, when required, instead of a permutation and vice-versa