If we have an expression \[^{n}{{C}_{12}}{{=}^{n}}{{C}_{6}}\], then \[^{n}{{C}_{2}}\]?
Answer
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Hint: We are given a question based on the combinations. We have an equality given to us using that we have to find \[^{n}{{C}_{2}}\]. We know if \[^{n}{{C}_{r}}{{=}^{n}}{{C}_{p}}\], then we can say that either \[r=p\] or we can say that \[r=n-p\]. Using this property, we will first find the value of the ‘n’. Then, we will substitute the value ‘n’ in \[^{n}{{C}_{2}}\]. Hence, we will have the value of \[^{n}{{C}_{2}}\].
Complete step-by-step solution:
According to the given question, we have to find the value of \[^{n}{{C}_{2}}\]. We are also given an equality condition, that is, \[^{n}{{C}_{12}}{{=}^{n}}{{C}_{6}}\]
We know about a certain property in combinations that, if \[^{n}{{C}_{r}}{{=}^{n}}{{C}_{p}}\], then we can say that either \[r=p\] or we can say that \[r=n-p\]. We will be solving the given question using this property.
We will first find the value of ‘n’ and then we will proceed to find the value of \[^{n}{{C}_{2}}\].
We are given that,
\[^{n}{{C}_{12}}{{=}^{n}}{{C}_{6}}\]
So, we will have \[r=n-p\], that is,
\[^{n}{{C}_{n-12}}{{=}^{n}}{{C}_{6}}\]
We can now write,
\[\Rightarrow n-12=6\]
We will write the expression in terms of ‘n’. we get,
\[\Rightarrow n=12+6\]
\[\Rightarrow n=18\]
So, we get the value of ‘n’ as 18.
Now, we will substitute this value of ‘n’ in the expression, \[^{n}{{C}_{2}}\].
We will use the formula for combinations to expand the expression, the formula is \[^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}\].
We have,
\[^{18}{{C}_{2}}=\dfrac{18!}{2!(18-2)!}\]
Solving the expression, we get,
\[{{\Rightarrow }^{18}}{{C}_{2}}=\dfrac{18!}{2!16!}\]
Expanding the factorial, we get,
\[{{\Rightarrow }^{18}}{{C}_{2}}=\dfrac{18\times 17\times 16!}{2!16!}\]
Cancelling the similar factorial, we get,
\[{{\Rightarrow }^{18}}{{C}_{2}}=\dfrac{18\times 17}{2\times 1}\]
\[{{\Rightarrow }^{18}}{{C}_{2}}=9\times 17\]
\[{{\Rightarrow }^{18}}{{C}_{2}}=153\]
Therefore, the value of \[^{18}{{C}_{2}}=153\].
Note: The calculation of the combination solution should be done step wise and that the factorials involved should be carefully written with (!) sign. The formula of the combinations should be correctly written as well.
Complete step-by-step solution:
According to the given question, we have to find the value of \[^{n}{{C}_{2}}\]. We are also given an equality condition, that is, \[^{n}{{C}_{12}}{{=}^{n}}{{C}_{6}}\]
We know about a certain property in combinations that, if \[^{n}{{C}_{r}}{{=}^{n}}{{C}_{p}}\], then we can say that either \[r=p\] or we can say that \[r=n-p\]. We will be solving the given question using this property.
We will first find the value of ‘n’ and then we will proceed to find the value of \[^{n}{{C}_{2}}\].
We are given that,
\[^{n}{{C}_{12}}{{=}^{n}}{{C}_{6}}\]
So, we will have \[r=n-p\], that is,
\[^{n}{{C}_{n-12}}{{=}^{n}}{{C}_{6}}\]
We can now write,
\[\Rightarrow n-12=6\]
We will write the expression in terms of ‘n’. we get,
\[\Rightarrow n=12+6\]
\[\Rightarrow n=18\]
So, we get the value of ‘n’ as 18.
Now, we will substitute this value of ‘n’ in the expression, \[^{n}{{C}_{2}}\].
We will use the formula for combinations to expand the expression, the formula is \[^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}\].
We have,
\[^{18}{{C}_{2}}=\dfrac{18!}{2!(18-2)!}\]
Solving the expression, we get,
\[{{\Rightarrow }^{18}}{{C}_{2}}=\dfrac{18!}{2!16!}\]
Expanding the factorial, we get,
\[{{\Rightarrow }^{18}}{{C}_{2}}=\dfrac{18\times 17\times 16!}{2!16!}\]
Cancelling the similar factorial, we get,
\[{{\Rightarrow }^{18}}{{C}_{2}}=\dfrac{18\times 17}{2\times 1}\]
\[{{\Rightarrow }^{18}}{{C}_{2}}=9\times 17\]
\[{{\Rightarrow }^{18}}{{C}_{2}}=153\]
Therefore, the value of \[^{18}{{C}_{2}}=153\].
Note: The calculation of the combination solution should be done step wise and that the factorials involved should be carefully written with (!) sign. The formula of the combinations should be correctly written as well.
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