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Chapter 6: Problem 18

How many strings of 20-decimal digits are there that contain two 0s, four 1s, three 2s, one 3, two 4s, three 5s, two 7s, and three 9s?

Short Answer

Expert verified
Using the multinomial coefficient, the number of strings is given by \( \frac{20!}{20736} \).

Step by step solution

01

Identify the Problem

Determine how many different strings of 20 decimal digits satisfy the condition of containing specific numbers of each digit.
02

Understand the Combination Formula

Use the multinomial coefficient to find the number of ways to arrange these digits. The multinomial coefficient is given by: \[ \frac{n!}{k_1! k_2! \, ... \, k_m!} \] where n is the total number of items to arrange, and each k represents the frequency of a specific item.
03

Apply the Multinomial Coefficient

In this problem, we have 20 digits in total and specific frequencies for each digit. Calculate the multinomial coefficient: \[ \frac{20!}{2! \times 4! \times 3! \times 1! \times 2! \times 3! \times 0! \times 2! \times 3!} \]
04

Compute the Factorial Values

First, compute the factorials:\[20! = 20 \times 19 \times 18 \times ... \times 1 \]\[2! = 2 \times 1 = 2\]\[4! = 4 \times 3 \times 2 \times 1 = 24\]\[3! = 3 \times 2 \times 1 = 6\]\[1! = 1\]\[0! = 1\]
05

Plug in the Values to the Formula

Insert the factorial values into the multinomial coefficient and simplify: \[ \frac{20!}{2! \times 4! \times 3! \times 1! \times 2! \times 3! \times 1! \times 2! \times 3!} = \frac{20!}{2 \times 24 \times 6 \times 1 \times 2 \times 6 \times 1 \times 2 \times 6}\]
06

Final Calculation

Now, perform the final calculation: \[ = \frac{20!}{20736}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is an area of mathematics focused on counting and arranging objects. It deals with problems where you need to figure out the number of ways to perform a task or arrange items with certain restrictions. Understanding combinatorics helps solve various problems involving selection, arrangement, and grouping elements.

In our exercise, combinatorics allows us to calculate how many distinct strings of 20 decimal digits we can form given specific frequencies for each digit. We rely on the multinomial coefficient for this calculation, providing a systematic way to account for the different arrangements of digits.
Factorials
Factorials are key in solving combinatorics problems. Written as n!, a factorial represents the product of all positive integers up to n. For example:
  • 3! = 3 脳 2 脳 1 = 6
  • 4! = 4 脳 3 脳 2 脳 1 = 24

The exercise involves calculating factorials of various numbers to determine the multinomial coefficient. Each digit in our string appears a certain number of times, and the factorial of those counts appears in the formula. Factorials generalize permutations where each unique arrangement of inputs is counted, contributing to the final solution in multinomial problems.
Permutations
Permutations refer to the different ways you can arrange a set of items. In simpler terms, if you have a sequence of elements, the number of permutations gives you all the possible ways to order those elements. This is deeply tied to factorials since the number of permutations of n unique items is n!.

In our problem, we deal with permutations of a set of digits with specific frequencies. Because some digits repeat, we use the multinomial coefficient to account for these repetitions. The multinomial formula adjusts the count of permutations by dividing by the factorials of the frequencies of each repeating element.
So if we arrange digits like '2, 2, 3,' we consider those repetitions by dividing the total permutations by the factorial of the number of times each digit repeats.

Using permutations in this controlled form helps us solve complex digit arrangement problems, as in this exercise, ensuring we count only the valid, unique strings as required.

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