Question

Each of 8 identical balls is to be placed in the squares shown in the figure given in a horizontal direction such that one horizontal row contains 6 balls and the other horizontal row contains 2 balls (Two balls can be in the same row or in different rows). In how many maximum different ways can this be done?

• A. 38
• B. 28
• C. 16
• D. 14

Answer 1

The correct option is A 38

The 6 balls must be on either of the middle rows. This can be done in 2 ways. Once, we put the 6 balls in their single horizontal row - it becomes evident that for placing the 2 remaining balls on a straight line there are 2 principal options:
1. Placing the two balls in one of the four rows with two squares. In this case, this numbers of ways of placing the balls in any particular row would be 1 way ( since once you were to choose one of the 4 rows, the balls would automatically get placed as there are only two squares in each row). Thus the total number of ways would be 2 $\times$ 4 $\times$1 = 8 ways.
2. Placing the two balls in the other row with six squares. In this case, the number of ways of placing the 2 balls in that row would be ${}^6{C}_2$. This would give us ${}^2{C}_1\times 1\times {}^6{C}_2$ = 30 ways. Total is 30 + 8 = 38 ways.