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.
Total balls in A box = 14, Total balls in A box = 12
A box = 1/2(8c1/14c1) = 2/7
B box = 1/2(6c1/12c1) = 1/4 —> total Probability = 2/7 + 1/4 =15/28
No. of ways = (4 x 3 x 2 x 1)/ (2 x 2) = 6 ways.
Total possible numbers with 6 at unit's place = 3; Total possible numbers with 3 at unit's place = 3
Total possible numbers with 5 at unit's place = 6
Thus 5 will come at unit's place 6 times, 3 will come 3 times and 6 will come 3 times. Sum of unit's column
= 6 × 3 + 3 × 3 + 5 × 6 = 57. This sum will be same for all column
Therefore
Sum will be 63327.
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 .