Candidates’ performance in this question was fair.  They performed better in the part (b)
than in the part (a).
In part (a),  ncr  + ncr-1 =    n!      +     n!                      =       n!(n-r+1)+n!(r)
                                              r!(n-r)   (r-1)!(n-r+1)!                        r!(n-r+1)!                      
                                   = n!(n+1)          =           (n+1)!      =     n+1Cr
                                      r!(n-r+1)!                   r!(n+1-r)

In part (b) probability of 2 obtaining distinction = 6C2 (0.04)2 (0.96)4  = 0.0204.
Probability of at most 3 of them obtaining distinction  =  1-p(more than three candidates obtaining distinction) =  1- (6C4 (0.04)4 (0.96)2 + 6C5 (0.04)5 (0.96) + 6C6 (0.04)6) =  1 – (0.000035389 + 0.000000589 + 0.000000004)  = 0.99996 approximately.