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Q22E

Expert-verifiedFound in: Page 414

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**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.

Letters \( = ABCDEFGH\).

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

**Formula:**

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

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.

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.

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.

\(GH\)

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.

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.

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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