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Chapter 1: Problem 9

A child has 12 blocks, of which 6 are black, 4 are red, 1 is white, and 1 is blue. If the child puts the blocks in a line, how many arrangements are possible?

Short Answer

Expert verified
The number of different arrangements for the 12 blocks with 6 black, 4 red, 1 white, and 1 blue is 27,720 using the formula for permutations with repetition: \(\frac{12!}{6! \times 4! \times 1! \times 1!}\).

Step by step solution

01

Identify the problem

This problem can be identified as a permutation problem that involves repetitions. We have 12 blocks in total, where 6 are black, 4 are red, 1 is white, and 1 is blue. We want to find the number of different arrangements of these blocks in a line.
02

Use the formula for permutations with repetition

In a permutation problem with repetition, we can use the following formula to calculate the number of arrangements: \(\frac{n!}{n_{1}! \times n_{2}! \times \cdots \times n_{k}!}\) Where: - n is the total number of elements (blocks in this case) - \(n_{1}\), \(n_{2}\), ..., \(n_{k}\) are the numbers of each type of element (number of each color of the blocks) In our case, n = 12 (total number of blocks), \(n_{1}\) = 6 (number of black blocks), \(n_{2}\) = 4 (number of red blocks), \(n_{3}\) = 1 (number of white blocks), and \(n_{4}\) = 1 (number of blue blocks).
03

Apply the formula

We will now apply the formula to find the number of arrangements: \(\frac{12!}{6! \times 4! \times 1! \times 1!}\) We need to calculate 12!, 6!, 4!, 1!, and 1!: \(12! = 12 \times 11 \times 10 \times \cdots \times 1 = 479001600\) \(6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720\) \(4! = 4 \times 3 \times 2 \times 1 = 24\) \(1! = 1\) \(1! = 1\) Now, we'll divide 12! by the product of 6!, 4!, 1!, and 1!: \(\frac{479001600}{(720 \times 24 \times 1 \times 1)}\) \(\frac{479001600}{17280}\)
04

Calculate the result

After dividing, we get the result: \(\frac{479001600}{17280} = 27720\) So there are 27,720 different possible arrangements of the 12 blocks in a line.

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