The correct option is A 91
We know that, nCr=n!r!(n−r)!
Given, 19Cr19Cr−1=23
=>19!r!(19−r)!19!(r−1)!(19−r+1)!=23
=>(r−1)!(20−r)!r!(19−r)!=23
=(r−1)!×(20−r)×(19−r)!r×(r−1)!×(19−r)!=23
=>20−rr=23
=>r=12
So,
14C12=14!12!(14−12)!=14!12!(2)!=14×13×12!12!×2=7×13=91