Question

There are 20 books on Algebra and Calculus in one library. For the greatest number of selections each of which consists of 5 books on each topic possible number of Algebra books are \[N\] then the value of \[\dfrac{N}{2}\] is.

Answer 1

Hint: Here, we will find the number of ways of selecting 5 Algebra and 5 Calculus books respectively. Then, we will find the total number of ways of selection and with the help of median, we will maximize the value of \[N\], hence, finding the required value of \[\dfrac{N}{2}\].

Formula Used:
We will use the following formulas:
1. \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] , where \[n\] is the total number of terms and \[r\] is the number of terms to be selected.
2. Median of odd terms \[ = \left( {\dfrac{{n + 1}}{2}} \right)\], where \[n\] is the total number of observations.

Complete step-by-step answer:
According to the question, in a library,
Total number of Algebra and Calculus books \[ = 20\]
Let the number of Algebra books be \[k\].
Therefore, number of Calculus books \[ = \left( {20 - k} \right)\]
Now, according to the question, the greatest number of selection of books on each topic is 5.
Hence, number of ways of selecting 5 books which are of Algebra \[ = {}^k{C_5}\]
Similarly, number of ways of selecting 5 books which are of Calculus \[ = {}^{20 - k}{C_5}\]
Therefore, total number of ways of selection \[ = {}^k{C_5} \times {}^{20 - k}{C_5}\]
Now, we know that \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
Hence, total number of ways of selection \[ = \dfrac{{k!}}{{5!\left( {k - 5} \right)!}} \times \dfrac{{\left( {20 - k} \right)!}}{{5!\left( {15 - k} \right)!}}\]
\[ \Rightarrow \]Total number of ways of selection \[ = \dfrac{{k\left( {k - 1} \right)\left( {k - 2} \right)\left( {k - 3} \right)\left( {k - 4} \right)}}{{5!}} \times \dfrac{{\left( {20 - k} \right)\left( {19 - k} \right)\left( {18 - k} \right)\left( {17 - k} \right)\left( {16 - k} \right)}}{{5!}}\]
Now, we have to maximize the numerator.
Here, for the values \[1,2,3,4,16,17,18,19,20\]; \[k = 0\]
Therefore, the remaining numbers which \[k\] can take value of are: \[5,6,7,8,9,10,11,12,13,14,15\].
Here, total number of terms \[ = 11\]
Now we will use the median formula to find the maximum value.
Median \[ = \left( {\dfrac{{n + 1}}{2}} \right) = \left( {\dfrac{{11 + 1}}{2}} \right)\]
\[ \Rightarrow \] Median \[ = \dfrac{{12}}{2} = {6^{th}}\]term
In this sequence, the sixth term is 10.
For, \[k = 10\], we will get the maximum value.

Therefore, possible number of algebra books \[ = k = N = 10\]
Hence, \[\dfrac{N}{2} = \dfrac{{10}}{2} = 5\]

Note:
Another way to solve this question is:
Total number of Algebra and Calculus books \[ = 20\]
Let the number of Algebra books be \[k\]
Therefore, number of Calculus books \[ = \left( {20 - k} \right)\]
Now, according to the question, the greatest number of selection of books on each topic is 5.
Hence, number of ways of selecting 5 books which are of Algebra \[ = {}^k{C_5}\]
Similarly, number of ways of selecting 5 books which are of Calculus \[ = {}^{20 - k}{C_5}\]
Therefore, total number of ways of selection \[ = {}^k{C_5} \times {}^{20 - k}{C_5}\]
Now, it is a fact that \[{}^n{C_r} \times {}^{N - n}{C_r}\] is maximum when \[n = \dfrac{N}{2}\].
Here, \[n = k\] and \[N = 20\]
Therefore, \[k = \dfrac{{20}}{2} = 10\]
Possible number of algebra books \[ = k = N = 10\]
Hence, \[\dfrac{N}{2} = \dfrac{{10}}{2} = 5\]