If n−1C3+n−1C4>nC3, then,
n≥4
n>5
n>7
none of these
We have,
n−1C3+π−1C4>nC3
= nC4>nC3 ∵nCr−1+nCr=n+1Cr
=n!(n−4)!4!>n!(n−3)!3!⇒14>1n−3⇒n>7