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

If it’s given that nC2=nC8^{n}{{C}_{2}}{{=}^{n}}{{C}_{8}}, we have to find nC2^{n}{{C}_{2}}.

Explanation

Solution

Hint: nCr^{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.
nCr=n!(n−r)!×r!^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}
Where n!n!or n factorial is, n! = n×(n−1)×(n−2)×(n−3)×….×3×2×1n!\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!=4×3×2×1=244!=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:
nCa=nCb^{n}{{C}_{a}}{{=}^{n}}{{C}_{b}} then either a=ba=b or a+b=na+b=n ……………………... (eq. 1)
Example:
5C2=5C3^{5}{{C}_{2}}{{=}^{5}}{{C}_{3}}
Proof:
We’ll try to prove it in the primitive method.
Consider LHS,
5C2=5!(5−2)!×2!=5!(3)!×2!^{5}{{C}_{2}}=\dfrac{5!}{\left( 5-2 \right)!\times 2!}=\dfrac{5!}{\left( 3 \right)!\times 2!} (LHS)
Consider RHS,
5C3=5!(5−2)!×3!=5!(2)!×3!^{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, a=2, b=3n=5,\text{ }a=2,\text{ }b=3
Now we can see it satisfies the relation n=a+bn=a+b as 5=2+35=2+3.
Thus we can use this result in a generalized way and solve the given question.
Explanation:
Step 1: It is given that
nC2=nC8^{n}{{C}_{2}}{{=}^{n}}{{C}_{8}}
and we know if nCa=nCb^{n}{{C}_{a}}{{=}^{n}}{{C}_{b}} either a=ba=b or n=a+bn=a+b,
Step 2: Here we can see that a≠b(2≠8)a\ne b(2\ne 8) so n=a+bn=a+b must be satisfied.
Therefore, n=a+b=2+8n=a+b=2+8 so we get
Step 3: Now, we have to find
10C2=10!(10−2)!×2!=10!(8)!×2!=10×9×8×7×6×5×4×3×2×18×7×6×5×4×3×2×1×2×1=45^{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 4545 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,
nC2=n!(n−2)!×2!{{n}_{C}}_{_{2}}=\dfrac{n!}{\left( n-2 \right)!\times 2!}
Consider RHS,
nC8=n!(n−8)!×8!^{n}{{C}_{8}}=\dfrac{n!}{\left( n-8 \right)!\times 8!}
Now as given in question we equate LHS and RHS,
n!(n−2)!×2!=n!(n−8)!×8!\dfrac{\text{n}!}{\left( \text{n}-2 \right)!\times 2!}=\dfrac{\text{n}!}{\left( \text{n}-8 \right)!\times 8!} we get,
2!×(n−2)!=8!×(n−8)!2!\times (n-2)!=8!\times (n-8)!
Now on comparing we see that n−2=8n-2=8 and n−8=2n-8=2 both of them imply that n=10n=10.