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

If nC12=nC8{}^{n}{{C}_{12}}={}^{n}{{C}_{8}} , then find the value of nC17{}^{n}{{C}_{17}}.

Explanation

Solution

From the given question first, we have to find the value of nn and then we have to find the nC17{}^{n}{{C}_{17}}. To find the value of nC17{}^{n}{{C}_{17}} we have to use the formula of nCr{}^{n}{{C}_{r}}.

Complete step by step solution:
We know that in the combination formula i.e. nCr{}^{n}{{C}_{r}}. we have a property that is
⇒nCr=nCn−r\Rightarrow {}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}
We can say that the value of nn is the sum of the values of rr and n−rn-r.
For suppose if
⇒nCx=nCy\Rightarrow {}^{n}{{C}_{x}}={}^{n}{{C}_{y}}
Then nn is equal to the sum of the xx and yy.
That is
⇒n=x+y\Rightarrow n=x+y
Here, from the question given that nC12=nC8{}^{n}{{C}_{12}}={}^{n}{{C}_{8}}.
Therefore, from the above property the value of nn is the sum of 1212 and 88.
the value of nn is
⇒n=12+8\Rightarrow n=12+8
⇒n=20\Rightarrow n=20
Now, we got the value of nnis 2020.
Now, we know that the formula of nCr{}^{n}{{C}_{r}}.
The formula is
⇒nCr=n!r!(n−r)!\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
Therefore, nC17{}^{n}{{C}_{17}}is 20C17{}^{20}{{C}_{17}}.
The expansion of 20C17{}^{20}{{C}_{17}} is done by the above formula
By, expanding we will get
⇒20C17=20!17!(20−17)!\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20!}{17!\left( 20-17 \right)!}
Here, ! means factorial
The formula of n!n! is written below
⇒n!=n×(n−1)×(n−2)…×3×2×1\Rightarrow n!=n\times \left( n-1 \right)\times \left( n-2 \right)\ldots \times 3\times 2\times 1
Therefore, 20!20!can be written as 20×19×18×17!20\times 19\times 18\times 17!
Then We can write 20C17{}^{20}{{C}_{17}} as
⇒20C17=20!17!(20−17)!\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20!}{17!\left( 20-17 \right)!}
⇒20C17=20×19×18×17!17!(20−17)!\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18\times 17!}{17!\left( 20-17 \right)!}
Here, 17!17! will be cancelled in both denominator and numerator
The remaining part is
⇒20C17=20×19×18(20−17)!\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18}{\left( 20-17 \right)!}
⇒20C17=20×19×183!\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18}{3!}
Here, 3!3! can be written as 3×2×13\times 2\times 1
⇒20C17=20×19×18(20−17)!\Rightarrow {}^{20}{{C}_{17}}=\dfrac{20\times 19\times 18}{\left( 20-17 \right)!}
⇒20C17=20×19×3\Rightarrow {}^{20}{{C}_{17}}=20\times 19\times 3
Therefore, the required answer is
⇒20C17=1140\Rightarrow {}^{20}{{C}_{17}}=1140

The value of nC17{}^{n}{{C}_{17}} is 11401140

Note: Students should know the basic formula and properties of nCr{}^{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. nCr{}^{n}{{C}_{r}} and nCp{}^{n}{{C}_{p}}. Students should know the basic difference between these two.
the formula of nCr{}^{n}{{C}_{r}} is
⇒nCr=n!r!(n−r)!\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
the formula of nCp{}^{n}{{C}_{p}} is
⇒nCp=n!(n−r)!\Rightarrow {}^{n}{{C}_{p}}=\dfrac{n!}{\left( n-r \right)!}