Permutation Test - 12

Out of 9 men, 2 men can be chosen in ⁹C₂ ways.

Since no husband and wife are to play in the same game, we have to select 2 women from the remaining 7 women. 

This can be done in ⁷C₂ ways. 

If M₁, M₂ and W₁, W₂ are chosen, then a team can be constituted in 2 ways, viz. M₁ W₁ or M₁ W₂. 

Thus, number of ways of arranging the game = ⁹C₂ x ⁷C₂ x 2 = 1512

Total number of persons = 3 girls + 9 boys = 12

Total number of seats = 2 × 3 + 4 × 2 = 14
So, total number of ways in which 12 persons can be seated on 14 seats = Number of arrangements of 14 seats by taking 12 at a time = ¹⁴P₁₂
Three girls can be seated together in the back row on adjacent seats in the following ways:
1, 2, 3 or 2, 3, 4 of first van
And 1, 2, 3 or 2, 3, 4 of second van
In each way, the three girls can interchange seats among themselves in 3! ways. 
So, total number of ways in which three girls can be seated together in the back row on adjacent seats = 4 × 3! 
Now, 9 boys are to be seated on remaining 11 seats, which can be done in ¹¹P₉ ways. 

Hence, by the fundamental principle of counting, total number of seating arrangements = ¹¹P₉ × 4 × 3!

Let an arrangement of digits of a number be x₁ x₂ x₃ x₄ x₅ x₆ x₇ x₈ x₉. Note that we require product of each of (x₁, x₂, x₃, x₄, x₅); (x₂, x₃, x₄, x₅, x₆); …; (x₅, x₆, x₇, x₈, x₉) to be divisible by 7.

This is possible if the 5^th digit is 7.

Therefore, we can arrange the 9 digits in desired number of ways,

Since position of 7 is fixed, the number of arrangements of remaining 8 digits is 8!.

We have 0 + 1 + 2 + 3 … + 8 + 9 = 45
To obtain an eight-digit number exactly divisible by 9, we must not use (0, 9) or (1, 8) or (2, 7) or (3, 6) or (4, 5). [Sum of the remaining eight digits is 36, which is exactly divisible by 9]
When we do not use (0, 9), then the number of required 8 digit numbers is 8!.
When one of (1, 8) or (2, 7) or (3, 6) or (4, 5) is not used, the remaining digits can be arranged in 8! - 7! ways.
{0 cannot be at extreme left} 
Hence, there are 8! + 4(8! - 7!) = (36)(7!) numbers in the desired category