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Q22E

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Discrete Mathematics and its Applications
Found in: Page 414
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

How many permutations of the letters \(ABCDEFGH\) contain

a) the string \(ED\)?

b) the string \(CDE\)?

c) the strings \(BA\) and \(FGH\)?

d) the strings \(AB\;,\;DE\) and \(GH\)?

e) the strings \(CAB\) and \(BED\)?

f) the strings \(BCA\) and \(ABF\)?

a) There are \(5040\) ways of permutation.

b) There are \(720\) ways of permutation.

c) There are \(120\) ways of permutation.

d) There are \(120\) ways of permutation.

e) There are \(24\) ways of permutation.

f) There is no permutation with both strings.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given data

Letters \( = ABCDEFGH\).

Step 2: Concept of Permutation

The word "permutation" refers to the act or process of changing the linear order of an ordered set.

Formula:

\(_n{P_r} = \frac{{n!}}{{(n - r)!}}\)

Step 3: Calculation for the permutations of the letters for string \(ED\)

a)

Assume the string as a single letter.

So, the letter becomes \(ABCDEFGH\).

Find the permutations of \(7\) letters, use the formula of permutations:

Number of permutations \( = P(7,7)\)

\(\begin{array}{l}P(7,7) = \frac{{7!}}{{(7 - 7)!}}\\P(7,7) = \frac{{7!}}{{0!}}\\P(7,7) = 5040\end{array}\)

Hence, there are \(5040\) ways of permutation.

Step 4: Calculation for the permutations of the letters for string \(CDE\)

b)

Assume the string as a single letter.

So, the letter becomes \(ABCDEFGH\).

Find the permutations of \(6\) letters, use the formula of permutations:

Number of permutations \( = P(6,6)\)

\(\begin{array}{l}P(6,6) = \frac{{6!}}{{(6 - 6)!}}\\P(6,6) = \frac{{6!}}{{0!}}\\P(6,6) = 720\end{array}\)

Hence, there are \(720\) ways of permutation.

Step 5: Calculation for the permutations of the letters for string \(BA\) and \(FGH\)

c)

Assume the string as a single letter.

So, the letter becomes \(ABCDEFGH\).

Find the permutations of \(5\) letters, use the formula of permutations:

Number of permutations \( = P(5,5)\)

\(\begin{array}{l}P(5,5) = \frac{{5!}}{{(5 - 5)!}}\\P(5,5) = \frac{{5!}}{{0!}}\\P(5,5) = 120\end{array}\)

Hence, there are \(120\) ways of permutation.

Step 6: Calculation for the permutations of the letters for string \(ABDE\) and \(GH\)

\(GH\)

d)

Assume the string as a single letter.

So, the letter becomes \(ABCDEFGH\).

Find the permutations of \(5\) letters, use the formula of permutations:

Number of permutations \( = P(5,5)\)

\(\begin{array}{l}P(5,5) = \frac{{5!}}{{(5 - 5)!}}\\P(5,5) = \frac{{5!}}{{0!}}\\P(5,5) = 120\end{array}\)

Hence, there are \(120\) ways of permutation.

Step 7: Calculation for the permutations of the letters for string \(AB\;,\;DE\) and \(GH\)

e)

Assume the string contains \(CABED\) as a single letter because \(B\) can occur only once.

So, the letter becomes \(CABEDFGH\).

Find the permutations of \(4\) letters, use the formula of permutations:

Number of permutations \( = P(4,4)\)

\(\begin{array}{l}P(4,4) = \frac{{4!}}{{(4 - 4)!}}\\P(4,4) = \frac{{4!}}{{0!}}\\P(4,4) = 24\end{array}\)

Hence, there are \(24\) ways of permutation.

Step 8: Calculation for the permutations of the letters for string \(BCA\) and \(ABF\)

f)

String cannot contain both \(BCA\) and \(ABF\) because the string \(B\) cannot be follow directly \(C\) and \(F\) both.

Hence, there is no permutation with both strings.

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