Question

How many strings of six lowercase letters from the English alphabet contain

a) the letter\(a\)?

b) the letters\(a\)and\(b\)?

c) the letters\(a\)and\(b\)in consecutive positions with\(a\)preceding\(b\), with all the letters distinct?

d) the letters\(a\)and\(b\), where\(a\)is somewhere to the left of\(b\)in the string, with all the letters distinct?

Short answer

1. The possible strings are\(64,775,151\).
2. The possible strings are\(11,737,502\).
3. The possible strings are\(1,275,120\).
4. The possible strings are\(3,825,360\).

Step-by-step solution

Definitions of Product rule, Permutation, and Combination

Product rule: If one event can occur in\(m\)ways and a second event can occur in\(n\)ways, then the number of ways that the two events can occur in sequence is then\(m \cdot n\).

Definition permutation (order is important):

\(P(n,r) = \frac{{n!}}{{(n - r)!}}\)

Definition combination (order is not important):

\(C(n,r) = \left( {\begin{array}{*{20}{l}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)with \(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\)

The Possible strings in total containing letter a.

(a)

The alphabet contains\(26\)letters.

For each of the\(6\)letters, there are\(26\)possible ways.

Use the product rule:

\(26 \cdot 26 \cdot 26 \cdot 26 \cdot 26 \cdot 26 = {26^6}\)

Possible strings not containing\(a\)

For each of the\(6\)letters, there are\(25\)possible ways (since \(a\) is not a possibility).

Use the product rule:

\(25 \cdot 25 \cdot 25 \cdot 25 \cdot 25 \cdot 25 = {25^6}\)Possible strings not containing \(a\)

The number of possible strings containing \(a\) is then the total number of possible strings decreased by the number of strings not containing \(a\):

\({26^6} - {25^6} = 64,775,151\)

Possible strings containing \(a\) and \(b\)

(b)

For each of the\(6\)letters, there are\(25\)possible ways since \(b\) is not a possibility.

Use the product rule:

\(25 \cdot 25 \cdot 25 \cdot 25 \cdot 25 \cdot 25 = {25^6}\)

For each of the\(6\)letters, there are\(25\)possible ways for strings not containing a nor b.

Use the product rule:

\(25 \cdot 25 \cdot 25 \cdot 25 \cdot 25 \cdot 25 = {25^6}\)

For each of the\(6\)letters, there are\(24\)possible ways since \(a\) and \(b\) both are not a possibility.

Use the product rule:

\(24 \cdot 24 \cdot 24 \cdot 24 \cdot 24 \cdot 24 = {24^6}\)

Use the subtract rule for the possible strings not containing \(a\) and \(b\).

\({25^6} + {25^6} - {24^6} = 2 \cdot {25^6} - {24^6}\)

The number of possible strings containing \(a\) and \(b\) is then the total number of possible strings decreased by the number of strings not containing \(a\) and \(b\) :

\(\begin{array}{l}{26^6} - \left( {2 \cdot {{25}^6} - {{24}^6}} \right) = {26^6} - 2 \cdot {25^6} + {24^6}\\{26^6} - \left( {2 \cdot {{25}^6} - {{24}^6}} \right) = 11,737,502\end{array}\)

Possible positions for \(ab\)

(c)

Since\(ab\)needs to have consecutive positions, the possible strings are of the form:

\(\begin{array}{l}ab{x_1}{x_2}{x_3}{x_4}\\{x_1}ab{x_2}{x_3}{x_4}\\{x_1}{x_2}ab{x_3}{x_4}\end{array}\)

\(\begin{array}{l}{x_1}{x_2}{x_3}ab{x_4}\\{x_1}{x_2}{x_3}{x_4}ab\end{array}\)

Thus, note that there are\(5\)positions for\(ab\)

Possible strings of the remaining letters:

The order of the letters matter (because a different order leads to a different string) and thus, use a permutation.

Since the letters are distinct, then select\(4\)letters from the remaining\(24\)letters.

\(\begin{array}{*{20}{l}}{n = 24}\\{r = 4}\end{array}\)

Use the definition of a permutation:

\(\begin{array}{l}P(24,4) = \frac{{24!}}{{(24 - 4)!}}\\P(24,4) = \frac{{24!}}{{20!}}\\P(24,4) = 24 \cdot 23 \cdot 22 \cdot 21\\P(24,4) = 255,024\end{array}\)

Possible strings containing\(ab\)and all different letters:

Use the product rule:

\(5 \cdot 255,024 = 1,275,120\)

Possible positions for\(a\)and\(b\)  where a is somewhere to the left of b

(d)

The order of the letters matter (because a different order leads to a different string) and thus, use a permutation.

Select 2 positions out of the 6 possible positions

Use the definition of a permutation:

\(\begin{array}{l}P(6,2) = \frac{{6!}}{{(6 - 2)!}}\\P(6,2) = \frac{{6!}}{{4!}}\\P(6,2) = 30\end{array}\)

Possible positions for \(a\) and \(b\) with \(a\) preceding \(b\).

Exactly half of the permutations will have \(a\) preceding \(b\) (while the other half will have \(b\) preceding \(a\)).

\(\frac{{30}}{2} = 15\)

Possible strings of the remaining letters:

The order of the letters matter (because a different order leads to a different string) and thus, use a permutation.

Since the letters are distinct, then select\(4\)letters from the remaining\(24\)letters.

\(\begin{array}{l}n = 24\\r = 4\end{array}\)

Use the definition of a permutation:

\(\begin{array}{l}P(24,4) = \frac{{24!}}{{(24 - 4)!}}\\P(24,4) = \frac{{24!}}{{20!}}\\P(24,4) = 24 \cdot 23 \cdot 22 \cdot 21\\P(24,4) = 255,024\end{array}\)

Possible strings containing \(a\) preceding \(b\) and all different letters:

Use the product rule:

\(15 \cdot 255,024 = 3,825,360\)