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Question

Mathematics Question on permutations and combinations

If n= mC2n= \,^mC_2, then the value of nC2^nC_2 is given by

A

3(m+1C4)3\left(^{m+1}C_{4}\right)

B

m−1C4^{m-1}C_{4}

C

m+1C4^{m+1}C_{4}

D

2(m+2C4)2\left(^{m+2}C_{4}\right)

Answer

3(m+1C4)3\left(^{m+1}C_{4}\right)

Explanation

Solution

n=mC2=m(m−1)2n=^{m}C_{2}=\frac{m\left(m-1\right)}{2} Since mm and (m−1)\left(m - 1\right) are two consecutive natural numbers, therefore their product is an even natural number. So m(m−1)2\frac{m\left(m-1\right)}{2} is also a natural number. Now m(m−1)2=m2−m2\frac{m\left(m-1\right)}{2}=\frac{m^{2}-m}{2} ∴m(m−1)2C2=(m2−m2)(m2−m2−1)2\therefore \frac{m\left(m-1\right)}{2}C_{2}=\frac{\left(\frac{m^{2}-m}{2}\right)\left(\frac{m^{2}-m}{2}-1\right)}{2} =m(m−1)(m2−m−2)8=\frac{m\left(m-1\right)\left(m^{2}-m-2\right)}{8} =m(m−1)[m2−2m+m−2]8=\frac{m\left(m-1\right)\left[m^{2}-2m+m-2\right]}{8} =m(m−1)[m(m2−2)+1(m−2)]8=\frac{m\left(m-1\right)\left[m\left(m^{2}-2\right)+1\left(m-2\right)\right]}{8} =m(m−1)[m(m−2)(m+1)]8=\frac{m\left(m-1\right)\left[m\left(m-2\right)\left(m+1\right)\right]}{8} =3×(m+1)m(m−1)(m−2)4×3×2×1=3(m+1C4)=\frac{3\times\left(m+1\right)m\left(m-1\right)\left(m-2\right)}{4\times3\times2\times1}=3\left(^{m+1}C_{4}\right)