Question

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

a) the string \(BCD\)?

b) the string \(CFGA\)?

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

d) the strings \(ABC\)and \(DE\)?

e) the strings \(ABC\)and \(CDE\)?

f) the strings \(CBA\)and \(BED\)?.

Short answer

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

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

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

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

e) There are \(6\) ways of permutation.

f) There is no permutation with both strings.

Step-by-step solution

Given data

Letters \( = ABCDEFG\)

Concept of Permutation

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

Formula:

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

Calculation for the permutations of the letters for string \(BCD\)

a)

Given string \( = BCD\)

Assume the string as a single letter.

So, the letter becomes \(ABCDEFG\).

We 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.

Calculation for the permutations of the letters for string \(CFGA\)

b)

Given string \(\; = CFGA\)

Assume the string as a single letter.

So, the letter becomes \(BDECFGA\).

We 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.

Calculation for the permutations of the letters for string  \(BA\) and \(GF\)

c)

Given string \( = BA\) and \(GF\)

Assume the string \(BA\) as a single letter and the string \(GF\) as a single letter.

So, the letter becomes \(ABCDEFG\).

We 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.

Calculation for the permutations of the letters for string \(ABC\)and \(DE\)

d)

Given string \( = ABC\) and \(DE\)

Assume the string \(ABC\) as a single letter and the string \(DE\) as a single letter.

So, the letter becomes \(ABCDEFG\).

We 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.

Calculation for the permutations of the letters for string \(ABC\)and \(CDE\)

e)

Given string \( = ABC\) and \(CDE\)

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

So, the letter becomes \(ABCDEFG\).

We find the permutations of \(3\) letters, use the formula of permutations.

Number of permutations \( = P(3,3)\)

\(\begin{array}{l}P(3,3) = \frac{{3!}}{{(3 - 3)!}}\\P(3,3) = \frac{{3!}}{{0!}}\\P(3,3) = 6\end{array}\)

Hence, there are \(6\) ways of permutation.

Calculation for the permutations of the letters for string \(\;CBA\)and \(\;BED\)

f)

Given string \( = CBA\) and \(BED\)

String cannot contain both \(CBA\) and \(BED\) because the string \(B\) cannot be follow directly \(A\) and \(E\) both.

Hence, there is no permutation with both strings.