A shelf contains $30 $ books in a row, how many ways there are to choose $4 $ books so between each pair of chosen books there are at least $2$ non chosen ones.
Tried to transform the question to stars and bars . But still have no clues. Any hints would be helpful .
On
Hint: Consider the alternate problem Arrange 26 balls(books) and 4 sticks(these are the books you are choosing. Slightly modify it since you need 2 books between all 4 chosen ones, so set 6 balls aside and put them in between the 3 slots between the 4 sticks in the beginning and now arrange the remaining 20 balls in 5 possible slots. This is the hint, to see the answer just move your cursor over the yellow space below.
Answer:
Spoiler textThis is the same as the problem- non-negative integral solutions of the equation $x_1+x_2+x_3+x_4+x_5=20$ which is just$\binom{24}{4}=21.22.23=10,626$
Each admissible choice can be encoded as a binary word of length $30$ containing exactly $4$ ones, whereby the first three ones have at least two zeros immediately following. Deleting these zeros gives a binary word of length $24$ with $4$ ones and no extra conditions. Conversely: Given any binary word of length $24$ containing $4$ ones insert two zeros after the first three ones, and you obtain an admissible selection of $4$ books from the shelf. The number $N$ you are looking for therefore is $$N={24\choose4}=10\,626\ .$$