Answer

Verified

382.2k+ 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, \[\dfrac{N}{2} = \dfrac{{10}}{2} = 5\]

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

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE