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Using the digits 2 through 8 , find the number of different 5 -digit numbers such that: (a) Digits cannot be repeated and must be written in increasing order. For example, 23678 is okay, but 32678 is not. (b) Digits can be repeated and must be written in non-decreasing order. For example, 24448 is okay, but 24484 is not.

Short Answer

Expert verified
For part (a): 21 different numbers can be created. For part (b): 462 different numbers can be created.

Step by step solution

01

Understanding the permutation without repetition

For part (a), we need to count the number of ways to choose 5 different digits out of 7 (digits 2 through 8) where the order matters because the digits must be arranged in an increasing sequence. Since the digits are in increasing order, each selection of 5 digits will create a unique number. This is a combination problem, not a permutation because once the five digits are chosen, they have only one possible order (increasing).
02

Calculating the combination for part (a)

Use the combination formula which is given by: \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \) where \( n \) is the total number of digits to choose from (which is 7 here) and \( k \) is the number of digits we want to choose (which is 5 here). Therefore, \( \binom{7}{5} = \frac{7!}{5!(7 - 5)!} \) simplifies to \( \binom{7}{5} = \frac{7 \times 6}{2 \times 1} = 21 \).
03

Understanding the permutation with repetition

For part (b), we are again selecting 5 digits from 7 choices (digits 2 through 8), but this time digits can be repeated and they must be in non-decreasing order. This is like distributing 5 identical items into 7 different bins. This is a problem of combinations with repetition.
04

Calculating the combination with repetition for part (b)

Use the formula for combinations with repetition, which is: \( \binom{n + k - 1}{k} \) where \( n \) is the number of types of items (here the digits), and \( k \) is the number of items to choose (digits in the number). Here, \( n = 7 \) and \( k = 5 \), so \( \binom{7 + 5 - 1}{5} = \binom{11}{5} \). Using the combination formula, \( \binom{11}{5} = \frac{11!}{5!6!} \) which simplifies to \( \binom{11}{5} = 462 \).

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Key Concepts

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

Permutation without Repetition
In discrete mathematics, permutation without repetition is used to count the number of ways items can be arranged when order matters and each item can only be used once. Considering our digits 2 through 8, when we create 5-digit numbers without repetition and in increasing order, we essentially perform a combination task, since the order is fixed once the digits are chosen.

Imagine you're lining up unique toys on a shelf where they must be in a set order by size; similarly, for the numbers to be in increasing order, there's only one way to arrange the chosen digits. The confusion between permutations and combinations arises because we're used to permutations allowing multiple orders, but the increasing order requirement makes our problem more of a combination type, despite being about the arrangement of numbers.
Combination Formula
The combination formula, denoted as \( \binom{n}{k} \), allows us to calculate the number of ways to choose a subset when the order doesn't matter. This is akin to selecting k flavors out of n available when making an ice cream sundae; you're interested in the flavors, not the order they're scooped. In part (a), where we choose 5 digits from the 7 available, we use this formula. Remember, the factorial notation (e.g., \( n! \) which stands for \( n \times (n-1) \times ... \times 1 \)) is essential when using this formula to evaluate the combination, reflecting the total number of arrangements divided by the arrangement counts of both the chosen set and the remainder.
Permutation with Repetition
Permutation with repetition is a concept where the order of items still matters, but items may be used more than once. In our context, 5-digit numbers can have repeated digits. Imagine you're allowed to pick the same flavor multiple times while preparing a five-scoop sundae. There are more possibilities than when each flavor was unique, as repetition increases the count of potential outcomes.

This concept translates into the problem where the sequence's non-decreasing order signifies that a digit's repetition doesn't alter the sequence's rank, much like how adding more of the same flavor doesn't affect the order of scoops in our sundae.
Combinations with Repetition
Combinations with repetition expand on the basic idea of combinations by allowing the same item to be selected more than once. This is like picking several pieces of fruit with the allowance to choose the same fruit multiple times; the order of selection does not matter. The formula for this, \( \binom{n + k - 1}{k} \), considers the repeatable items and adjusts the combination count accordingly.

For the number example in part (b), the formula reflects that each digit position can be any of the seven digits, much like each fruit pick can be any available fruit, regardless of previous picks. The result is a much higher count compared to when repetitions were not allowed, as the opportunity for repetition increases the overall selection possibilities.

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Most popular questions from this chapter

A multiset is a collection of objects, just like a set, but can contain an object more than once (the order of the elements still doesn't matter). For example, \\{1,1,2,5,5,7\\} is a multiset of size 6 . (a) How many sets of size 5 can be made using the 10 numeric digits 0 through \(9 ?\) (b) How many multisets of size 5 can be made using the 10 numeric digits 0 through \(9 ?\)

Ten ladies of a certain age drop off their red hats at the hat check of a museum. As they are leaving, the hat check attendant gives the hats back randomly. In how many ways can exactly six of the ladies receive their own hat (and the other four not)? Explain.

Consider all 5 letter "words" made from the letters \(a\) through \(h\). (Recall, words are just strings of letters, not necessarily actual English words.) (a) How many of these words are there total? (b) How many of these words contain no repeated letters? (c) How many of these words start with the sub-word "aha"? (d) How many of these words either start with "aha" or end with "bah" or both? (e) How many of the words containing no repeats also do not contain the sub- word "bad"?

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Let \(A=\\{1,2,3,4,5\\}\). How many injective functions \(f: A \rightarrow A\) have the property that for each \(x \in A, f(x) \neq x ?\)

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