Question

Boxes numbered 1, 2, 3, 4, and 5 are kept in a row, and they are necessary to be filled with either a red ball or a blue ball such that no two adjacent boxes can be filled with blue balls. How many different arrangements are possible, given that the balls of a given colour are exactly identical in all respects?
(a) 8
(b) 10
(c) 13
(d) 22

Answer 1

Hint: If you think a bit you will notice that there can be a maximum of 3 blue balls so that they can be arranged according to the condition given in the question. So, you will have three cases: 5 red balls, 1 blue and 4 red balls, 2 blue balls and 3 red balls, 3 blue and 2 red balls. Solve and find the number of arrangements satisfying the conditions in each case separately and add them to get the answer.

Complete step by step solution:
Let us start by making cases and finding the number of ways of executing the case. First, let us say that all the balls are red, and the number of ways of executing this case is 1. Now let us take the case that there are 4 red balls and one blue ball. So, the number of ways of executing this case is 5, i.e., out of the 5 dustbins, one will contain the blue ball.
Now let us check for the case of 2 blue balls and three red balls. So, the number of ways of executing this is equal to the number of ways of selecting 2 dustbins out of five, i.e., $^{5}{{C}_{2}}=10$ . But out of these 10 possibilities, we need to subtract those when both blue balls are in adjacent boxes, which is equal to 4, i.e., blue balls in box 1 and 2 or in box 2 and 3 or in box 3 and 4 or in box 4 and 5. Therefore, the number of possible cases such that 2 blue balls are present and are not in adjacent boxes is 6.
Now when we take 3 blue balls, there is only 1 possibility, i.e., blue balls are kept in box 1, box 3, and box 5. Also, we cannot take more than 3 blue balls, as more than 3 blue balls cannot be arranged in a manner that none of them are adjacent.
Therefore, total number of possibilities = 1 + 5 + 6 + 1 = 13.
Therefore, the answer to the above question is option (c).

Note: The main point in the above solution where you might get confused is in the case where we are taking 2 blue balls and 3 red balls. In this case you should be very careful and subtract those cases in which the blue balls are appearing in adjacent boxes. It is suggested that you don’t try to find the number of cases when two same coloured balls are not together rather you find all possible ways of putting the balls and subtract those cases when same coloured balls are together to get the answer, to avoid mistakes.