   Chapter 9.3, Problem 38ES ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

#### Solutions

Chapter
Section ### Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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# For each of exerciser 37-39, the number of dements in a certain set can be found by computing the number in a larger universe that are not in the set and subtracting this from the total in the larger universe. In each of these, as was the case for the solution to Example 9.3.6<b). De Morgan's laws and the inclusion/ exclusion rule can be used. How many permutations of abcde are there in which the first character is a, b or c and the but character is c, d. or e?

To determine

To find:

Permutations of abcde that are there in which the first character is a, b or c and the last character is c, d or e.

Explanation

Given information:

The number of elements ina certain set can be found by computing the number in somelarger universe that are not in the set and subtracting this fromthe total.

Calculation:

We know that:

N(AB)=N(U)N( ( AB )c)                     (Difference law)=N(U)N(AcBc)                       (De Morgan's law)  =N(U)(N( A c )+N( B c )N( A c B c ))           (Inclusion/Exclusion rule for two sets) Let

A = Permutations of abcde in which the first character is a, b or c B = Permutations of abcde in which the second character is c, d or e

In general, we know that there are 5! Permutations of abcde, since abcde contains 5 elements.

N(U)=5!=5×4×3×2×1=210

We then note that we are interested in the same number of elements in AB, which contains all permutations of abcde in which the first character is a, b or c and the last character is c, d or e.

Number of permutations of abcde in which the first character is not a, not b and not c :

A permutation is an ordering of the letters in abcde, which contains five letters

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