Question

In how many ways can a dozen books be placed on four distinguishable shelves

a) if the books are indistinguishable copies of the same title?

b) if no two books are the same, and the positions of the books on the shelves matter? (Hint: Break this into 12 tasks, placing each book separately. Start with the sequence 1, 2, 3, 4 to represent the shelves. Represent the books by bi , i = 1, 2, . . . , 12. Place b1 to the right of one of the terms in 1, 2, 3, 4. Then successively place b2, b3, . . . , and b12.)

Short answer

1. Therefore, there are \(455\)different ways
2. Therefore, there are \(217945728000\) different ways if no two books are the same, and the positions of the books on the shelves matter

Step-by-step solution

Step 1: If books are indistinguishable copies of same title

There are four shelves with twelve indistinguishable books, so we use combination and repetition to get the number of ways we can order the books.

\(\begin{array}{l}C_{4 - 1}^{12 + 4 - 1} \to C_3^{15}\\C_r^n = \frac{{n!}}{{r!(n - r)!}} = \frac{{15!}}{{3!12!}}\\C_3^{15} = \frac{{2730}}{6} = 455\end{array}\)

Therefore, there are \(455\)different ways

Step 2: if no two books are the same, and the positions of the books on the shelves matter

There are four ways we can place book 1, five ways for book and since the order matters six ways for book 3 and fifteen ways for book 12.

Multiply those together to get the answer\(\begin{array}{l}4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12 \times 13 \times 14 \times 15\\ = 217945728000\end{array}\)

Therefore, there are \(217945728000\) different ways if no two books are the same, and the positions of the books on the shelves matter