
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
446.4k+ views
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, \[\dfrac{N}{2} = \dfrac{{10}}{2} = 5\]
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, \[\dfrac{N}{2} = \dfrac{{10}}{2} = 5\]
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
