Ten books on a shelf are arranged randomly. Find the…
Ten books on a shelf are arranged randomly. Find the probability that in such an arrangement four specific books will be together.
Explanation of Method
All orders of the ten books are equally likely, so we count permutations. The trick for counting orders in which four books stand together is to glue them into one block: then we arrange $7$ objects (the block plus $6$ other books), and multiply by the number of ways to order the four books inside the block. Dividing by $10!$ gives the probability.
Step 1: All outcomes
An outcome is an order of the $10$ books on the shelf. The number of orders of $10$ different objects is
\[ 10! = 10\cdot9\cdot8\cdots2\cdot1 = 3\,628\,800, \]
and all of them are equally likely.
Step 2: Favourable outcomes — the block method
- the 4 specific books, glued into one block
- the other 6 books
- Glue the four specific books into one block. Now there are $6 + 1 = 7$ objects to arrange: $7!$ ways.
- Inside the block the four books can stand in any order: $4!$ ways.
Every arrangement of the $7$ objects combines with every order inside the block:
\[ \text{favourable} = 7!\cdot 4! = 5040\cdot 24 = 120\,960. \]
Step 3: The probability
\[ P = \frac{7!\,4!}{10!} = \frac{4!}{10\cdot9\cdot8} = \frac{24}{720}, \]
\[ \boxed{P = \frac{1}{30} \approx 0.033.} \]
Step 4: Check by counting positions
Look only at which places the four books occupy. There are $\binom{10}{4}=210$ equally likely sets of places. The places are consecutive in $7$ cases (places $1$–$4$, $2$–$5$, …, $7$–$10$). So
\[ P = \frac{7}{210} = \frac{1}{30}. \ \checkmark \]
Step 5: General formula
For $n$ books of which $k$ specific ones must stand together:
\[ P = \frac{(n-k+1)!\,k!}{n!} = \frac{n-k+1}{\binom{n}{k}}. \]
Result
| Answer | |
|---|---|
| All orders | $10!$ |
| Favourable | $7!\cdot4!$ (block of 4 + 6 books, times order inside the block) |
| Probability | $P=\dfrac{7!\,4!}{10!}=\dfrac{1}{30}\approx0.033$ |