The correct option is C 142
5 letters can go into 5 envelopes in 5! ways, i.e. 5! = 5×4×3×2×1× = 20×6= 120
and there is just 1 combination in which each and every letter goes to its corresponding envelope, thus 120−1=119 for first condition
1 letter in its correct corresponding envelope and 4 others juggled, here we get 4! = = 4×3×2×1× = 24 again we should subtract 1 combination in which all 4 letters can go correctly to their respective envelopes, thus 24−1 = 23 for Second condition.
= 119+23=142 ways.