Solve the given expression of permutation and combination, the expression is:
\[^{2n}{C_2} - {(^2}{C_2}) = \]?
A. \[{n^2}\]
B. \[{(n - 1)^2}\]
C. \[{(n + 1)^2}\]
D. \[2{n^2} - n - 1\]
Answer
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Hint: Here the given question belongs to permutation and combination, here the expression is of combination, and to solve this expression here we need to use the property of combination and get the simplified answer. The formulae of combination is:
\[{ = ^n}{C_m} = \dfrac{{n!}}{{r!(n - r)!}}\]
Complete step-by-step answer:
Here to solve the given expression we need to use the above formulae of combination to get the simplified answer:
On solving we get:
\[
= \dfrac{{2n!}}{{2!(2n - 2)!}} - \dfrac{{2!}}{{2!(2 - 2)!}} \\
= \dfrac{{2n \times (2n - 1) \times (2n - 2)!}}{{2!(2n - 2)!}} - \dfrac{{2!}}{{2! \times 0!}} \\
= \dfrac{{2n(2n - 1)}}{{2 \times 1}} - 1 \\
= 2{n^2} - n - 1 \\
\]
Here we get the final simplified answer for the expression.
So, the correct answer is “Option D”.
Note: Permutation and combination is the set of rule in mathematics to solve for the given set of problems, here these rules need to be used exactly in order to get the exact solution. Every formulae and relation is predefined to solve further.
\[{ = ^n}{C_m} = \dfrac{{n!}}{{r!(n - r)!}}\]
Complete step-by-step answer:
Here to solve the given expression we need to use the above formulae of combination to get the simplified answer:
On solving we get:
\[
= \dfrac{{2n!}}{{2!(2n - 2)!}} - \dfrac{{2!}}{{2!(2 - 2)!}} \\
= \dfrac{{2n \times (2n - 1) \times (2n - 2)!}}{{2!(2n - 2)!}} - \dfrac{{2!}}{{2! \times 0!}} \\
= \dfrac{{2n(2n - 1)}}{{2 \times 1}} - 1 \\
= 2{n^2} - n - 1 \\
\]
Here we get the final simplified answer for the expression.
So, the correct answer is “Option D”.
Note: Permutation and combination is the set of rule in mathematics to solve for the given set of problems, here these rules need to be used exactly in order to get the exact solution. Every formulae and relation is predefined to solve further.
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