A) \[3{{(}^{m+1}}{{C}_{4}})\]
B) \[^{m-1}{{C}_{4}}\]
C) \[^{m+1}{{C}_{4}}\]
D) \[2{{(}^{m+2}}{{C}_{4}})\]
Correct Answer: A
Solution :
\[n{{=}^{m}}{{C}_{2}}=\frac{m(m-1)}{2}\] Since m and (m - 1) are two consecutive natural numbers, therefore their product is an even natural number. So\[\frac{m(m-1)}{2}\] is also a natural number. Now\[\frac{m(m-1)}{2}=\frac{{{m}^{2}}-m}{2}\] \[\therefore \]\[\frac{m(m-1)}{2}{{C}_{2}}=\frac{\left( \frac{{{m}^{2}}-m}{2} \right)\left( \frac{{{m}^{2}}-m}{2}-1 \right)}{2}\] \[=\frac{m(m-1)({{m}^{2}}-m-2)}{8}\] \[=\frac{m(m-1)[{{m}^{2}}-2m+m-2]}{8}\] \[=\frac{m(m-1)[m(m-2)+1(m-2)]}{8}\] \[=\frac{m(m-1)(m-2)(m+1)}{8}\] \[=\frac{3\times (m+1)m(m-2)(m-2)}{4\times 3\times 2\times 1}=3\left( ^{m+1}{{C}_{4}} \right)\]
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