Question

If it’s given that \[^{n}{{C}_{2}}{{=}^{n}}{{C}_{8}}\], we have to find \[^{n}{{C}_{2}}\].

Answer 1

Hint: \[^{n}{{C}_{r}}\] is the formula to calculate combinations where n represents the number of items and r represents the number of items being chosen at a time.
\[^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}\]
Where \[n!\]or n factorial is, \[n!\text{ }=\text{ }n\times \left( n-1 \right)\times \left( n-2 \right)\times \left( n-3 \right)\times \ldots .\times 3\times 2\times 1\].
E.g. \[4!=4\times 3\times 2\times 1=24\].

Complete step-by-step solution -
There are various properties of this formula and the property used in this question is given below:
\[^{n}{{C}_{a}}{{=}^{n}}{{C}_{b}}\] then either \[a=b\] or \[a+b=n\] ……………………... (eq. 1)
Example:
\[^{5}{{C}_{2}}{{=}^{5}}{{C}_{3}}\]
Proof:
We’ll try to prove it in the primitive method.
Consider LHS,
\[^{5}{{C}_{2}}=\dfrac{5!}{\left( 5-2 \right)!\times 2!}=\dfrac{5!}{\left( 3 \right)!\times 2!}\] (LHS)
Consider RHS,
\[^{5}{{C}_{3}}=\dfrac{5!}{\left( 5-2 \right)!\times 3!}=\dfrac{5!}{\left( 2 \right)!\times 3!}\] (RHS)
We get LHS = RHS.
Also comparing it with equation 1 we know,
\[n=5,\text{ }a=2,\text{ }b=3\]
Now we can see it satisfies the relation \[n=a+b\] as \[5=2+3\].
Thus we can use this result in a generalized way and solve the given question.
Explanation:
Step 1: It is given that
\[^{n}{{C}_{2}}{{=}^{n}}{{C}_{8}}\]
and we know if \[^{n}{{C}_{a}}{{=}^{n}}{{C}_{b}}\] either \[a=b\] or \[n=a+b\],
Step 2: Here we can see that \[a\ne b(2\ne 8)\] so \[n=a+b\] must be satisfied.
Therefore, \[n=a+b=2+8\] so we get
Step 3: Now, we have to find
 $^{10}{{C}_{2}}=\dfrac{10!}{\left( 10-2 \right)!\times 2!}=\dfrac{10!}{\left( 8 \right)!\times 2!}=\dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\times 2\times 1}=45$
Hence, we obtain $45$ as the answer.

Note: Alternative method
We can also use primitive method to solve this problem i.e. directly substituting LHS and RHS in formula,
Consider LHS,
${{n}_{C}}_{_{2}}=\dfrac{n!}{\left( n-2 \right)!\times 2!}$
Consider RHS,
$^{n}{{C}_{8}}=\dfrac{n!}{\left( n-8 \right)!\times 8!}$
Now as given in question we equate LHS and RHS,
$\dfrac{\text{n}!}{\left( \text{n}-2 \right)!\times 2!}=\dfrac{\text{n}!}{\left( \text{n}-8 \right)!\times 8!}$ we get,
\[2!\times (n-2)!=8!\times (n-8)!\]
Now on comparing we see that \[n-2=8\] and \[n-8=2\] both of them imply that \[n=10\].