Question

Assertion: When the number of ways of arranging the 21 objects of which \[r\] objects are identical of one type and remaining are identical of second type is maximum, then the maximum value of \[{}^{13}{{C}_{r}}\] is 78
Reason: \[{}^{2n+1}{{C}_{r}}\] is maximum when \[r=n\]
(a) Both assertion and reason are correct and reason is correct explanation of assertion
(b) Both assertion and reason are correct but reason is not correct explanation of assertion
(c) Assertion is correct and reason is wrong
(d) Assertion is wrong and reason is correct

Answer 1

Hint: We solve this problem by using some standard formulas of combinations and arrangements,
We have the formula that \[{}^{n}{{C}_{r}}\] is maximum based on \[n\] is even or odd
When \[n\] is even then \[{}^{n}{{C}_{r}}\] is maximum if \[r=\dfrac{n}{2}\]
When \[n\] is odd then \[{}^{n}{{C}_{r}}\] is maximum if \[r=\dfrac{n-1}{2}\]
We also have the formula that the number of arrangements of \[n\] objects in which \[p\] are identical of one type and \[q\] are identical of other type is given as \[\dfrac{n!}{p!q!}\]

Complete step by step answer:
Let us take the reason first
We are given in the reason that the value \[{}^{2n+1}{{C}_{r}}\] is maximum when \[r=n\]
Let us assume that the given value as
\[\Rightarrow A={}^{2n+1}{{C}_{r}}\]
We know that the formula that \[{}^{n}{{C}_{r}}\] is maximum based on \[n\] is even or odd
When \[n\] is even then \[{}^{n}{{C}_{r}}\] is maximum if \[r=\dfrac{n}{2}\]
When \[n\] is odd then \[{}^{n}{{C}_{r}}\] is maximum if \[r=\dfrac{n-1}{2}\]
Here we can see that number \[2n+1\] in A is an odd number
By using the above formula we get that the value of \[r\] for which \[{}^{2n+1}{{C}_{r}}\] is maximum as
\[\begin{align}
  & \Rightarrow r=\dfrac{\left( 2n+1 \right)-1}{2} \\
 & \Rightarrow r=n \\
\end{align}\]
Therefore we can conclude that the reason is correct.
Now, let us check the assertion.
We are given that when the number of ways of arranging the 21 objects of which \[r\] objects are identical of one type and remaining are identical of second type is maximum, then the maximum value of \[{}^{13}{{C}_{r}}\] is 78
Let us assume that the number of arrangements of given 21 objects as \[N\]
Here, we can see that there are 21 objects in which \[r\] objects are identical of one type and \[21-r\] objects are identical of other type.
We know that the formula that the number of arrangements of \[n\] objects in which \[p\] are identical of one type and \[q\] are identical of other type is given as \[\dfrac{n!}{p!q!}\]
By using the above formula we get the number of arrangements of 21 objects as
\[\Rightarrow N=\dfrac{21!}{r!\left( 21-r \right)!}\]
We know that the formula of combinations that is
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the abominations formula in above equation we get
\[\begin{align}
  & \Rightarrow N={}^{21}{{C}_{r}} \\
 & \Rightarrow N={}^{2\left( 10 \right)+1}{{C}_{r}} \\
\end{align}\]
We are given that these arrangements are maximum.
We know that \[{}^{2n+1}{{C}_{r}}\] is maximum when \[r=n\]
By using the above condition we get the value of \[r\] as
\[\Rightarrow r=10\]
Now, let us assume that the value of \[{}^{13}{{C}_{r}}\] as \[x\]
By using the combinations formula we get the value of \[x\] as
\[\begin{align}
  & \Rightarrow x={}^{13}{{C}_{10}} \\
 & \Rightarrow x=\dfrac{13!}{10!\left( 13-10 \right)!} \\
 & \Rightarrow x=\dfrac{13\times 12\times 11\times 10!}{10!\times 3\times 2}=286 \\
\end{align}\]
We are given that the value of \[{}^{13}{{C}_{r}}\] as 78
Therefore, we can conclude that the assertion is wrong.
So, we have the answer as assertion is wrong and reason is correct
Therefore, option (d) is correct answer.


Note:
Students may do mistake in the number of arrangements in the assertion.
We have the formula that the number of arrangements of \[n\] objects in which \[p\] are identical of one type and \[q\] are identical of other type is given as \[\dfrac{n!}{p!q!}\]
But students may miss the point of identical objects and take the number of arrangements as \[n!\]
This is wrong because arranging the identical objects gives the same order but there will be change in the position of arrangements
So, we take the formula as \[\dfrac{n!}{p!q!}\]