We get 3 equations: x + y + z = 300, x + 2y + 5z = 960, 2x + y + 5z = 920.
Subtract 1) from 2) and 3) to get:
3x + 3y + 10z = 1880 and 3x + 3y + 3z = 900; 7z = 980 Hence z = 140.
No. of ways = (4 x 3 x 2 x 1)/ (2 x 2) = 6 ways.
Ways of having all two wrong out of 2 letters & 2 directed employees
ways of having all 3 wrong out of 3 letters to 3 directed envelops
= 3! – (1 (when all 3 are correct) + 3C1 × 1 (when one is correct and 2 one wrong)
= 6 – (1 + 3) = 2
ways of having all 4 wrong out of 4 latters and 4 envelops
= 4! – (1 + 4C1 × 2 (when one is correct and 3 wrong) + 4C2 × 1 (When 2 are correct and two are wrong)
= 4! – (1 + 8 + 6) = 9
ways of having all 5 wrong out of 5 letters and 5 envelops
= 5! – (1 + 5C1 × 9 + 5C2 × 2 + 5C3 × 1)
= 5! – (1 + 45 + 20 + 10)
= 44 .
