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# 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.

Last updated date: 20th Jul 2024
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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.

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, $\dfrac{N}{2} = \dfrac{{10}}{2} = 5$