Question

If \[{}^{n}{{C}_{12}}={}^{n}{{C}_{8}}\] , then find the value of \[{}^{n}{{C}_{17}}\].

Answer 1

Hint: From the given question first, we have to find the value of \[n\] and then we have to find the \[{}^{n}{{C}_{17}}\]. To find the value of \[{}^{n}{{C}_{17}}\] we have to use the formula of \[{}^{n}{{C}_{r}}\].

Complete step by step solution:
We know that in the combination formula i.e. \[{}^{n}{{C}_{r}}\]. we have a property that is
\[\Rightarrow {}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}\]
We can say that the value of \[n\] is the sum of the values of \[r\] and \[n-r\].
For suppose if
\[\Rightarrow {}^{n}{{C}_{x}}={}^{n}{{C}_{y}}\]
Then \[n\] is equal to the sum of the \[x\] and \[y\].
That is
\[\Rightarrow n=x+y\]
Here, from the question given that \[{}^{n}{{C}_{12}}={}^{n}{{C}_{8}}\].
Therefore, from the above property the value of \[n\] is the sum of \[12\] and \[8\].
the value of \[n\] is
\[\Rightarrow n=12+8\]
\[\Rightarrow n=20\]
Now, we got the value of \[n\]is \[20\].
Now, we know that the formula of \[{}^{n}{{C}_{r}}\].
The formula is
\[\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Therefore, \[{}^{n}{{C}_{17}}\]is \[{}^{20}{{C}_{17}}\].
The expansion of \[{}^{20}{{C}_{17}}\] is done by the above formula
By, expanding we will get
\[\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20!}{17!\left( 20-17 \right)!}\]
Here, ! means factorial
The formula of \[n!\] is written below
\[\Rightarrow n!=n\times \left( n-1 \right)\times \left( n-2 \right)\ldots \times 3\times 2\times 1\]
Therefore, \[20!\]can be written as \[20\times 19\times 18\times 17!\]
Then We can write \[{}^{20}{{C}_{17}}\] as
\[\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20!}{17!\left( 20-17 \right)!}\]
\[\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18\times 17!}{17!\left( 20-17 \right)!}\]
Here, \[17!\] will be cancelled in both denominator and numerator
The remaining part is
\[\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18}{\left( 20-17 \right)!}\]
\[\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18}{3!}\]
Here, \[3!\] can be written as \[3\times 2\times 1\]
\[\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18}{\left( 20-17 \right)!}\]
\[\Rightarrow {}^{20}{{C}_{17}}=20\times 19\times 3\]
Therefore, the required answer is
\[\Rightarrow {}^{20}{{C}_{17}}=1140\]

The value of \[{}^{n}{{C}_{17}}\] is \[1140\]

Note: Students should know the basic formula and properties of \[{}^{n}{{C}_{r}}\]. We should be very careful while doing the calculation and expanding the factorials. Student should not confuse between the formulas of combination and permutation i.e. \[{}^{n}{{C}_{r}}\] and \[{}^{n}{{C}_{p}}\]. Students should know the basic difference between these two.
the formula of \[{}^{n}{{C}_{r}}\] is
\[\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
the formula of \[{}^{n}{{C}_{p}}\] is
\[\Rightarrow {}^{n}{{C}_{p}}=\dfrac{n!}{\left( n-r \right)!}\]