If n= mC2, then the value of nC2 is given by

Question

If n= mC2, then the value of nC2 is given by (A) 3(m+1C4) (B) m-1C4 (C) m+1C4 (D) 2(m+2C4)

Tardigrade Admin · Accepted Answer

n=mC2=(m(m-1)/2) Since m and (m - 1) are two consecutive natural numbers, therefore their product is an even natural number. So (m(m-1)/2) is also a natural number. Now (m(m-1)/2)=(m2-m/2) ∴ (m(m-1)/2)C2=(((m2-m/2))((m2-m/2)-1)/2) =(m(m-1)(m2-m-2)/8) =(m(m-1)[m2-2m+m-2]/8) =(m(m-1)[m(m2-2)+1(m-2)]/8) =(m(m-1)[m(m-2)(m+1)]/8) =(3×(m+1)m(m-1)(m-2)/4×3×2×1)=3(m+1C4)

Q. If $n =^{m} C_{2}$ , then the value of $^{n} C_{2}$ is given by

$3 (^{m + 1} C_{4})$

$^{m - 1} C_{4}$

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

$2 (^{m + 2} C_{4})$

Q. If n=mC2​, then the value of nC2​ is given by

3(m+1C4​)

m−1C4​

m+1C4​

2(m+2C4​)

Q. If $n =^{m} C_{2}$ , then the value of $^{n} C_{2}$ is given by

$3 (^{m + 1} C_{4})$

$^{m - 1} C_{4}$

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

$2 (^{m + 2} C_{4})$