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# If ${}^n{C_4} = {}^n{C_6}$ , find ${}^{12}{C_n}$.

Last updated date: 20th Jun 2024
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Hint: We will use the property of the combinations which states that If we are given a condition that ${}^n{C_p} = {}^n{C_q}$, then any of these two situations will follow: (i): p = q and (ii): p + q = n to solve the given question. Here, n is the total number of ways to do a certain action then p or q are the individual ways to do it in a specific manner.

We are given a condition that ${}^n{C_4} = {}^n{C_6}$.
Then we know that we have a property of combinations which states that if ${}^n{C_p} = {}^n{C_q}$, then either p = q or p + q = n.
Here, we have ${}^n{C_4} = {}^n{C_6}$, we can say that here p = 4 and q = 6.
Using the above property in the given condition, we get
Either of the two conditions satisfy i. e.,
(i): p = q
$\Rightarrow 4$should be equal to 6 but $4 \ne 6$.
Hence, we can say that this situation is not possible.
Looking at the first situation, we get that second situation must follow as the first is not possible.
(ii): p + q = n
$\Rightarrow 4 + 6 = n \\ \Rightarrow 10 = n \\$
Hence, substituting this value of n in ${}^{12}{C_n}$, we get
$\Rightarrow {}^{12}{C_{10}}$ is the required value we need to calculate.
Upon simplification using the formula ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ , we get
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12!}}{{\left( {12 - 10} \right)!10!}}$
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12!}}{{2!10!}}$
$\Rightarrow {}^{12}{C_{10}} = \dfrac{{12 \times 11}}{{2 \times 1}} = 66$
Therefore, we can say that the value $\Rightarrow {}^{12}{C_{10}}$is 66.

Note: This problem is not tough but tricky. We can also find the required value by evaluating the given condition ${}^n{C_4} = {}^n{C_6}$ for the value of n and then substituting it in $\Rightarrow {}^{12}{C_{10}}$.