Question

Assertion: When the number of ways of arranging 21 objects of which r objects are identical of one type and remaining is identical of the 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 the reason is the correct explanation for the assertion.
(b) Both assertion and reason are correct and the reason is not the correct explanation for the assertion.
(c) Assertion is correct but the reason is incorrect.
(d) Assertion is incorrect but the reason is correct.

Answer 1

First, before proceeding for this, we must know that concept of combination that when ${}^{n}{{C}_{r}}$having n odd gives the maximum value when $r=\dfrac{n+1}{2}$ or$r=\dfrac{n-1}{2}$.Then, by using the concept mentioned above, we get the condition of the maximum for ${}^{2n+1}{{C}_{r}}$only when n is replaced by (2n+1) in the above value of r. Then, by applying the same concept that for the odd value of n which is 21 now gives the maximum only when $r=\dfrac{n+1}{2}$ or$r=\dfrac{n-1}{2}$, we get the maximum value for assertion and we can comment on correctness on assertion and reason.

Complete step-by-step solution
In this question, we are supposed to find whether the given statements are correct or not in which one is an assertion and the other is the reason for that.
So, before proceeding for this, we must know that concept of combination that when ${}^{n}{{C}_{r}}$having n odd gives the maximum value when $r=\dfrac{n+1}{2}$ or$r=\dfrac{n-1}{2}$.
Now, by using the concept mentioned above, we get the condition of the maximum for ${}^{2n+1}{{C}_{r}}$only when n is replaced by (2n+1) in the above value of r as:
$\begin{aligned} & r=\dfrac{2n+1+1}{2} \\\ & \Rightarrow r=\dfrac{2n+2}{2} \\\ & \Rightarrow r=n+1 \\\ \end{aligned}$ or $\begin{aligned} & r=\dfrac{2n+1-1}{2} \\\ & \Rightarrow r=\dfrac{2n}{2} \\\ & \Rightarrow r=n \\\ \end{aligned}$
So, we get the condition that ${}^{2n+1}{{C}_{r}}$gives maximum value when r=n as true which means reason is correct.
Now, checking for the assertion, we use the same concept for getting the maximum value points from 21 objects, r being selected which are identical gives the combination as:
${}^{21}{{C}_{r}}$
Then, by applying the same concept that for odd values of n which is 21 now gives the maximum only when $r=\dfrac{n+1}{2}$ or$r=\dfrac{n-1}{2}$.
Then, by substituting the value of n as 21 in the above two formulas, we get:
$\begin{aligned} & r=\dfrac{21+1}{2} \\\ & \Rightarrow r=\dfrac{22}{2} \\\ & \Rightarrow r=11 \\\ \end{aligned}$ or $\begin{aligned} & r=\dfrac{21-1}{2} \\\ & \Rightarrow r=\dfrac{20}{2} \\\ & \Rightarrow r=10 \\\ \end{aligned}$
So, we get two values of r where ${}^{21}{{C}_{r}}$gives the maximum value.
Now, we need to check which value of r whether 10 or 11 will give the maximum value when substituted in ${}^{13}{{C}_{r}}$to get the maximum value.
So, firstly we will check for the value of r as 10 in${}^{13}{{C}_{r}}$, we get:

& {}^{13}{{C}_{10}}=\dfrac{13!}{10!\left( 13-10 \right)!} \\\ & \Rightarrow {}^{13}{{C}_{10}}=\dfrac{13\times 12\times 11\times 10!}{10!\times 3\times 2\times 1} \\\ & \Rightarrow {}^{13}{{C}_{10}}=\dfrac{13\times 12\times 11}{3\times 2} \\\ & \Rightarrow {}^{13}{{C}_{10}}=13\times 2\times 11 \\\ & \Rightarrow {}^{13}{{C}_{10}}=286 \\\ \end{aligned}$$ Similarly, we find the value of r as 11 in${}^{13}{{C}_{r}}$, we get: $$\begin{aligned} & {}^{13}{{C}_{11}}=\dfrac{13!}{11!\left( 13-11 \right)!} \\\ & \Rightarrow {}^{13}{{C}_{11}}=\dfrac{13\times 12\times 11!}{11!\times 2\times 1} \\\ & \Rightarrow {}^{13}{{C}_{11}}=\dfrac{13\times 12}{2} \\\ & \Rightarrow {}^{13}{{C}_{11}}=13\times 6 \\\ & \Rightarrow {}^{13}{{C}_{11}}=78 \\\ \end{aligned}$$ So, we can see that we get the maximum value of ${}^{13}{{C}_{r}}$as 286 for the value of r as 10. **So, we get the statement in assertion as incorrect. Hence, option (d) is correct.** **Note:** Now, to solve these type of the questions we need to know some of the basics to get the value of ${}^{n}{{C}_{r}}$ is given by the formula of the combination: $^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ To find the factorial of the number n, multiply the number n with (n-1) till it reaches 1. To understand let us find the factorial of 4. $\begin{aligned} & 4!=4\times 3\times 2\times 1 \\\ & \Rightarrow 24 \\\ \end{aligned}$