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Found in: Page 414

### Discrete Mathematics and its Applications

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

# How many permutations of the letters $$ABCDEFGH$$ containa) 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 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.