Question

A bag contains \[n\] white and \[n\] black balls. Pairs of balls are drawn at random without replacement successively until the bag is empty. If the number of ways in which each pair consists of one white and one black ball is 14,400, then \[n\] is equal to
A.6
B.5
C.4
D.3

Answer 1

Hint: We are required to find the number of white balls or the black balls that are denoted by \[n\]. To solve this question, we will use the concept of the permutation and the combination, and the multiplication law of permutation and combinations. We will find out the combination for each pair and then we will add the total based on it.

Formula Used: We will use the formula for the combinations to solve this question,
\[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step-by-step answer:
We are given that a bag contains \[n\] white and \[n\] black balls, and pairs of balls are drawn at random without replacement successively until the bag is empty. We are required to find the value of \[n\]if the number of ways in which each pair consists of one white and one black ball is 14,400.
So, let us find the combination for the first pair –
Now we have \[n\]options of white balls for the first white ball. We are required to choose only one ball. So, we get the combination as \[{}^n{C_1}\].
Total number of combinations for first white ball \[ = {}^n{C_1}\]
Now we have \[n\]options of black balls for the first black ball. We are required to choose only one ball. So, we get the combination as \[{}^n{C_1}\].
Total number of combinations for first black ball \[ = {}^n{C_1}\]
Total number of combinations for the first pair \[ = {}^n{C_1} \cdot {}^n{C_1} = \dfrac{{n!}}{{1!\left( {n - 1} \right)!}} \cdot \dfrac{{n!}}{{1!\left( {n - 1} \right)!}}\]
Applying the factorial, we get
\[{}^n{C_1} \cdot {}^n{C_1} = n \cdot n = {n^2}\]
We will now find the combination for the second pair.
Now we have \[n - 1\] options of white balls for the second white ball. We are required to choose only one ball. So, we get the combination as \[{}^{n - 1}{C_1}\].
Total number of combinations for second white ball \[ = {}^{n - 1}{C_1}\]
Now we have \[n - 1\] options of black balls for the second black ball. We are required to choose only one ball. So, we get the combination as \[{}^{n - 1}{C_1}\].
Total number of combinations for second black ball \[ = {}^{n - 1}{C_1}\]
Total number of combinations for the second pair \[ = {}^{n - 1}{C_1} \cdot {}^{n - 1}{C_1} = \dfrac{{\left( {n - 1} \right)!}}{{1!\left\{ {\left( {n - 1} \right) - 1} \right\}!}} \cdot \dfrac{{\left( {n - 1} \right)!}}{{1!\left\{ {\left( {n - 1} \right) - 1} \right\}!}}\]
Solving the factorial, we get
\[\begin{array}{l}{}^{n - 1}{C_1} \cdot {}^{n - 1}{C_1} = \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 2} \right)!}} \cdot \dfrac{{\left( {n - 1} \right)!}}{{\left( {n - 2} \right)!}}\\ = \left( {n - 1} \right)\left( {n - 1} \right)\\ = {\left( {n - 1} \right)^2}\end{array}\]
Now we will find the combination for the last pair.
Now we have 1 option of the white ball for the last white ball. We are required to choose only one ball. So, we get the combination as \[{}^1{C_1}\].
Total number of combinations for the last white ball\[ = {}^1{C_1}\]
Now we have 1 option of the black ball for the last black ball. We are required to choose only one ball. So, we get the combination as \[{}^1{C_1}\].
Total number of combinations for the last black ball \[ = {}^1{C_1}\]
Total number of combinations for the last pair \[ = {}^1{C_1} \cdot {}^1{C_1} = \dfrac{{1!}}{{1!\left( {1 - 1} \right)!}} \cdot \dfrac{{1!}}{{1!\left( {1 - 1} \right)!}} = {1^2}\]
Now, according to the multiplication law of combinations,
Total number of combinations for all the pairs is \[{n^2} \cdot {\left( {n - 1} \right)^2} \cdot ..... \cdot {1^2}\]
Now we are given that the total number of pairs are 14,400. So, we get
\[\begin{array}{l}{n^2} \cdot {\left( {n - 1} \right)^2} \cdot ..... \cdot {1^2} = 14400\\ \Rightarrow {n^2} \cdot {\left( {n - 1} \right)^2} \cdot ..... \cdot {1^2} = {120^2}\end{array}\]
On taking under root on both sides of the equation, we get,
\[n \cdot \left( {n - 1} \right) \cdot ..... \cdot 1 = 120\]
Now we know that the LHS of the equation is the formula for the \[n!\]. We also know that 120 is equal to \[5!\]. This is because, \[n! = 1 \times 2 \times ..... \times \left( {n - 1} \right) \times n\].
So,
\[\begin{array}{l}1 \times 2 \times 3 \times 4 \times 5 = 120\\ = 5!\end{array}\]
On substituting these values in the above equation, and canceling out the factorial, we get,
\[\begin{array}{c}n! = 5!\\n = 5\end{array}\]
Hence, the correct answer is option (B).

Note: The multiplication law of combination states that if we can perform one job in \[a\] number of ways and another job in \[b\] number of ways, then we can perform both the jobs in \[a \times b\] number of ways.
Also, we can break down the formula for \[{}^n{C_1}\] as shown below.
\[\begin{array}{l}{}^n{C_1} = \dfrac{{n!}}{{1!\left( {n - 1} \right)!}}\\ = \dfrac{{n!}}{{\left( {n - 1} \right)!}}\\ = \dfrac{{n\left( {n - 1} \right)!}}{{\left( {n - 1} \right)!}}\\ = n\end{array}\]
Hence, we can see that when \[r = 1\] then the value of the combination is equal to the value of \[n\].