Question

If $^n{C_{10}}{ = ^n}{C_{15}}$, find $^{27}{C_n}$.

Answer 1

Hint: Here we can use the property, $^n{C_r}{ = ^n}{C_{n - r}}$. Using this property, we can easily find the value of $n$. After finding the value of $n$, we can use this to get the value of $^{27}{C_n}$.

Complete Step by Step Solution:
As we know,$^n{C_r}{ = ^n}{C_{n - r}}................$(equation $1$)
Given: $^n{C_{10}}{ = ^n}{C_{15}}.....................$(equation $2$)
Relating these two, we can conclude that $r = 10$.
Now, equation $2$will become:
$^n{C_{10}}{ = ^n}{C_{n - 10}}{ = ^n}{C_{15}}$
$
  \therefore n - 10 = 15 \\
   \Rightarrow n = 15 + 10 = 25 \\
 $
Now, we got the value of $n$. So, the value of $^{27}{C_n}$ can be calculated easily.
$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Using this formula,
$^{27}{C_n}$= $^{27}{C_{25}}$= $\dfrac{{27!}}{{(27 - 25)!25!}}$=$\dfrac{{27 \times 26 \times 25!}}{{2! \times 25!}} = \dfrac{{27 \times 26}}{2} = 351$ ($\because n! = [n(n - 1)](n - 2)!$)

Hence, the value of $^{27}{C_n}$ will be $351$.

Additional information:
$^n{C_r}$ represents the selection of objects from a group of objects where order of objects does not matter.
$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Where, $n$is the total number of objects and $r$is the number of selected objects.

Note:
Remember the properties and you will be able to solve such questions easily. In this question, we used only one property but others are equally important. You may need them in another question. Also, this question can be solved in other methods also. However, that is quite and the above mentioned method is the easiest one.
Method $2$: given, $^n{C_{10}}{ = ^n}{C_{15}}$
$ \Rightarrow \dfrac{{n!}}{{(n - 10)!10!}} = \dfrac{{n!}}{{(n - 15)!15!}}$
Solve this to get the value of $n$. And then substitute this value in $^{27}{C_n}$ to get its value.